Solving $log_5 (2x+1)=log_3 (3x-3)$.












1












$begingroup$


I am trying to resolve the equation $$log_5 (2x+1) = log_3 (3x-3)$$ and then of sketch the functions $y=log_5 (2x+1)$ and $y=log_3 (3x-3)$ get the solution $x=2$. There is an method that do not use the graphic method? Thanks for your suggestions. Any ideas will be appreciated.






logarithms







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$endgroup$








  • 2




    $begingroup$
    $$log_3(3x-3)cdotlog_5(3)=log_5(3x-3)$$
    $endgroup$
    – Don Thousand
    Dec 20 '18 at 19:10






  • 2




    $begingroup$
    You can hope that your teacher isn't out to torture you and $(2x+1) = 5^n$ and $(3x-3)= 3^n$ for some natural number $n.$ Start with, $0$, then try $1$ and find that a solution exists. After you find one solution, argue that both the LHS and RHS are upward sloping, but one is always more upward sloping than the other, so there can be no other intersection points.
    $endgroup$
    – Doug M
    Dec 20 '18 at 19:17








  • 1




    $begingroup$
    Hint: $$log_a(x)=frac{log_b(x)}{log_b(a)}.$$ (Well, that's where I'd start anyway . . . )
    $endgroup$
    – Shaun
    Dec 20 '18 at 19:18


















1












$begingroup$


I am trying to resolve the equation $$log_5 (2x+1) = log_3 (3x-3)$$ and then of sketch the functions $y=log_5 (2x+1)$ and $y=log_3 (3x-3)$ get the solution $x=2$. There is an method that do not use the graphic method? Thanks for your suggestions. Any ideas will be appreciated.






logarithms







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $$log_3(3x-3)cdotlog_5(3)=log_5(3x-3)$$
    $endgroup$
    – Don Thousand
    Dec 20 '18 at 19:10






  • 2




    $begingroup$
    You can hope that your teacher isn't out to torture you and $(2x+1) = 5^n$ and $(3x-3)= 3^n$ for some natural number $n.$ Start with, $0$, then try $1$ and find that a solution exists. After you find one solution, argue that both the LHS and RHS are upward sloping, but one is always more upward sloping than the other, so there can be no other intersection points.
    $endgroup$
    – Doug M
    Dec 20 '18 at 19:17








  • 1




    $begingroup$
    Hint: $$log_a(x)=frac{log_b(x)}{log_b(a)}.$$ (Well, that's where I'd start anyway . . . )
    $endgroup$
    – Shaun
    Dec 20 '18 at 19:18
















1












1








1





$begingroup$


I am trying to resolve the equation $$log_5 (2x+1) = log_3 (3x-3)$$ and then of sketch the functions $y=log_5 (2x+1)$ and $y=log_3 (3x-3)$ get the solution $x=2$. There is an method that do not use the graphic method? Thanks for your suggestions. Any ideas will be appreciated.






logarithms







share|cite|improve this question











$endgroup$




I am trying to resolve the equation $$log_5 (2x+1) = log_3 (3x-3)$$ and then of sketch the functions $y=log_5 (2x+1)$ and $y=log_3 (3x-3)$ get the solution $x=2$. There is an method that do not use the graphic method? Thanks for your suggestions. Any ideas will be appreciated.






logarithms




logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 19:41









Shaun

10.5k113687




10.5k113687










asked Dec 20 '18 at 19:01









Ricardo LargaespadaRicardo Largaespada

563213




563213








  • 2




    $begingroup$
    $$log_3(3x-3)cdotlog_5(3)=log_5(3x-3)$$
    $endgroup$
    – Don Thousand
    Dec 20 '18 at 19:10






  • 2




    $begingroup$
    You can hope that your teacher isn't out to torture you and $(2x+1) = 5^n$ and $(3x-3)= 3^n$ for some natural number $n.$ Start with, $0$, then try $1$ and find that a solution exists. After you find one solution, argue that both the LHS and RHS are upward sloping, but one is always more upward sloping than the other, so there can be no other intersection points.
    $endgroup$
    – Doug M
    Dec 20 '18 at 19:17








  • 1




    $begingroup$
    Hint: $$log_a(x)=frac{log_b(x)}{log_b(a)}.$$ (Well, that's where I'd start anyway . . . )
    $endgroup$
    – Shaun
    Dec 20 '18 at 19:18
















  • 2




    $begingroup$
    $$log_3(3x-3)cdotlog_5(3)=log_5(3x-3)$$
    $endgroup$
    – Don Thousand
    Dec 20 '18 at 19:10






  • 2




    $begingroup$
    You can hope that your teacher isn't out to torture you and $(2x+1) = 5^n$ and $(3x-3)= 3^n$ for some natural number $n.$ Start with, $0$, then try $1$ and find that a solution exists. After you find one solution, argue that both the LHS and RHS are upward sloping, but one is always more upward sloping than the other, so there can be no other intersection points.
    $endgroup$
    – Doug M
    Dec 20 '18 at 19:17








  • 1




    $begingroup$
    Hint: $$log_a(x)=frac{log_b(x)}{log_b(a)}.$$ (Well, that's where I'd start anyway . . . )
    $endgroup$
    – Shaun
    Dec 20 '18 at 19:18










2




2




$begingroup$
$$log_3(3x-3)cdotlog_5(3)=log_5(3x-3)$$
$endgroup$
– Don Thousand
Dec 20 '18 at 19:10




$begingroup$
$$log_3(3x-3)cdotlog_5(3)=log_5(3x-3)$$
$endgroup$
– Don Thousand
Dec 20 '18 at 19:10




2




2




$begingroup$
You can hope that your teacher isn't out to torture you and $(2x+1) = 5^n$ and $(3x-3)= 3^n$ for some natural number $n.$ Start with, $0$, then try $1$ and find that a solution exists. After you find one solution, argue that both the LHS and RHS are upward sloping, but one is always more upward sloping than the other, so there can be no other intersection points.
$endgroup$
– Doug M
Dec 20 '18 at 19:17






$begingroup$
You can hope that your teacher isn't out to torture you and $(2x+1) = 5^n$ and $(3x-3)= 3^n$ for some natural number $n.$ Start with, $0$, then try $1$ and find that a solution exists. After you find one solution, argue that both the LHS and RHS are upward sloping, but one is always more upward sloping than the other, so there can be no other intersection points.
$endgroup$
– Doug M
Dec 20 '18 at 19:17






1




1




$begingroup$
Hint: $$log_a(x)=frac{log_b(x)}{log_b(a)}.$$ (Well, that's where I'd start anyway . . . )
$endgroup$
– Shaun
Dec 20 '18 at 19:18






$begingroup$
Hint: $$log_a(x)=frac{log_b(x)}{log_b(a)}.$$ (Well, that's where I'd start anyway . . . )
$endgroup$
– Shaun
Dec 20 '18 at 19:18












2 Answers
2






active

oldest

votes


















2












$begingroup$

Let $$f(x) = log_5 (2x+1)-log_3(3x-3)$$ which is defined for $x>1$.



Since for $x>1$ we have $$(x+{1over 2})ln 5 > (x-1)ln5 >(x-1)ln3>0$$



we have also that $$f'(x) = {1over (x+{1over 2})ln 5 } - {1over (x-1)ln3} <0$$



so given function has at most one zero ...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is at most we can do at such equation. We are left to guess a solution somehow. Any way you can also try with some aproximaton method.
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 19:32










  • $begingroup$
    You right I already think about derivate the function $f(x)$ and verify that is monotonous. Then it has at most one zero.
    $endgroup$
    – Ricardo Largaespada
    Dec 20 '18 at 19:37



















0












$begingroup$

If you want a purely algebraic solution, although rather tedious, you can solve the equation as follows:



$$log_5(2x+1) = frac{log_3(2x+1)}{log_3 5} tag{1}$$



$$frac{log_3(2x+1)}{log_3 5} = log_3(3x-3) tag{2}$$



$$left(3^{log_3(2x+1)}right)^{frac{1}{log_3 5}} = 3x-3 tag{3}$$



$$(2x+1)^{frac{1}{log_3 5}} = 3x-3 tag{4}$$



$$(2x+1)^{log_5 3} = 3x-3 tag{5}$$



$$log_{2x+1} (3x-3) = log_5 3 tag{6}$$



$$boxed{x = 2} tag{7}$$



$(1)$: Change of bases: $log_a b = frac{log_c b}{log_c a}$



$(2)$: Rewriting the equation



$(3)$: Rewriting as an exponential: $c = log_a b iff a^c = b$



$(4)$: Simplifying



$(5)$: Inversion of base and argument: $frac{1}{log_a b} = log_b a$



$(6)$: Rewriting as a logarithm: $a^b = c iff log_a b = c$



$(7)$: Solution to $2x+1 = 5$ and $3x-3 = 3$



But of course, you have to check for any extra solutions. Since $log_a b^c = clog_a b$ and $log_{a^c} b = frac{1}{c}log_a b$, you have $$log_{a^ c} b^c = log_a b$$



$$implies log_5 3 = log_{5^n} 3^n$$



$$implieslog_{2x+1} (3x-3) = log_{5^n} 3^n$$



$$implies 2x+1 = 5^n; quad 3x-3 = 3^n$$



$$3x-3 = 3^n implies x-1 = 3^{n-1} implies x = 3^{n-1}+1$$



Plugging this in the first equation, you have



$$2left(3^{n-1}+1right) = 5^n implies 2left(3^{n-1}right)+3 = 5^n$$



It’s fairly obvious the only solution is $n = 1$ (since the RHS grows at a much faster rate), for which we already solved the equation. Hence, it’s the only solution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You are claim that $log_a b=log_A B leftrightarrow a=A cap b=B$.
    $endgroup$
    – Ricardo Largaespada
    Jan 1 at 7:25










  • $begingroup$
    If $A = a^n$ and $B = b^n$, then yes, $log_a b = log_A B$. (Of course, assuming $b > 0$.)
    $endgroup$
    – KM101
    Jan 1 at 7:28












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Let $$f(x) = log_5 (2x+1)-log_3(3x-3)$$ which is defined for $x>1$.



Since for $x>1$ we have $$(x+{1over 2})ln 5 > (x-1)ln5 >(x-1)ln3>0$$



we have also that $$f'(x) = {1over (x+{1over 2})ln 5 } - {1over (x-1)ln3} <0$$



so given function has at most one zero ...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is at most we can do at such equation. We are left to guess a solution somehow. Any way you can also try with some aproximaton method.
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 19:32










  • $begingroup$
    You right I already think about derivate the function $f(x)$ and verify that is monotonous. Then it has at most one zero.
    $endgroup$
    – Ricardo Largaespada
    Dec 20 '18 at 19:37
















2












$begingroup$

Let $$f(x) = log_5 (2x+1)-log_3(3x-3)$$ which is defined for $x>1$.



Since for $x>1$ we have $$(x+{1over 2})ln 5 > (x-1)ln5 >(x-1)ln3>0$$



we have also that $$f'(x) = {1over (x+{1over 2})ln 5 } - {1over (x-1)ln3} <0$$



so given function has at most one zero ...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is at most we can do at such equation. We are left to guess a solution somehow. Any way you can also try with some aproximaton method.
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 19:32










  • $begingroup$
    You right I already think about derivate the function $f(x)$ and verify that is monotonous. Then it has at most one zero.
    $endgroup$
    – Ricardo Largaespada
    Dec 20 '18 at 19:37














2












2








2





$begingroup$

Let $$f(x) = log_5 (2x+1)-log_3(3x-3)$$ which is defined for $x>1$.



Since for $x>1$ we have $$(x+{1over 2})ln 5 > (x-1)ln5 >(x-1)ln3>0$$



we have also that $$f'(x) = {1over (x+{1over 2})ln 5 } - {1over (x-1)ln3} <0$$



so given function has at most one zero ...






share|cite|improve this answer











$endgroup$



Let $$f(x) = log_5 (2x+1)-log_3(3x-3)$$ which is defined for $x>1$.



Since for $x>1$ we have $$(x+{1over 2})ln 5 > (x-1)ln5 >(x-1)ln3>0$$



we have also that $$f'(x) = {1over (x+{1over 2})ln 5 } - {1over (x-1)ln3} <0$$



so given function has at most one zero ...







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 19:38

























answered Dec 20 '18 at 19:22









Maria MazurMaria Mazur

49.9k1361125




49.9k1361125












  • $begingroup$
    That is at most we can do at such equation. We are left to guess a solution somehow. Any way you can also try with some aproximaton method.
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 19:32










  • $begingroup$
    You right I already think about derivate the function $f(x)$ and verify that is monotonous. Then it has at most one zero.
    $endgroup$
    – Ricardo Largaespada
    Dec 20 '18 at 19:37


















  • $begingroup$
    That is at most we can do at such equation. We are left to guess a solution somehow. Any way you can also try with some aproximaton method.
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 19:32










  • $begingroup$
    You right I already think about derivate the function $f(x)$ and verify that is monotonous. Then it has at most one zero.
    $endgroup$
    – Ricardo Largaespada
    Dec 20 '18 at 19:37
















$begingroup$
That is at most we can do at such equation. We are left to guess a solution somehow. Any way you can also try with some aproximaton method.
$endgroup$
– Maria Mazur
Dec 20 '18 at 19:32




$begingroup$
That is at most we can do at such equation. We are left to guess a solution somehow. Any way you can also try with some aproximaton method.
$endgroup$
– Maria Mazur
Dec 20 '18 at 19:32












$begingroup$
You right I already think about derivate the function $f(x)$ and verify that is monotonous. Then it has at most one zero.
$endgroup$
– Ricardo Largaespada
Dec 20 '18 at 19:37




$begingroup$
You right I already think about derivate the function $f(x)$ and verify that is monotonous. Then it has at most one zero.
$endgroup$
– Ricardo Largaespada
Dec 20 '18 at 19:37











0












$begingroup$

If you want a purely algebraic solution, although rather tedious, you can solve the equation as follows:



$$log_5(2x+1) = frac{log_3(2x+1)}{log_3 5} tag{1}$$



$$frac{log_3(2x+1)}{log_3 5} = log_3(3x-3) tag{2}$$



$$left(3^{log_3(2x+1)}right)^{frac{1}{log_3 5}} = 3x-3 tag{3}$$



$$(2x+1)^{frac{1}{log_3 5}} = 3x-3 tag{4}$$



$$(2x+1)^{log_5 3} = 3x-3 tag{5}$$



$$log_{2x+1} (3x-3) = log_5 3 tag{6}$$



$$boxed{x = 2} tag{7}$$



$(1)$: Change of bases: $log_a b = frac{log_c b}{log_c a}$



$(2)$: Rewriting the equation



$(3)$: Rewriting as an exponential: $c = log_a b iff a^c = b$



$(4)$: Simplifying



$(5)$: Inversion of base and argument: $frac{1}{log_a b} = log_b a$



$(6)$: Rewriting as a logarithm: $a^b = c iff log_a b = c$



$(7)$: Solution to $2x+1 = 5$ and $3x-3 = 3$



But of course, you have to check for any extra solutions. Since $log_a b^c = clog_a b$ and $log_{a^c} b = frac{1}{c}log_a b$, you have $$log_{a^ c} b^c = log_a b$$



$$implies log_5 3 = log_{5^n} 3^n$$



$$implieslog_{2x+1} (3x-3) = log_{5^n} 3^n$$



$$implies 2x+1 = 5^n; quad 3x-3 = 3^n$$



$$3x-3 = 3^n implies x-1 = 3^{n-1} implies x = 3^{n-1}+1$$



Plugging this in the first equation, you have



$$2left(3^{n-1}+1right) = 5^n implies 2left(3^{n-1}right)+3 = 5^n$$



It’s fairly obvious the only solution is $n = 1$ (since the RHS grows at a much faster rate), for which we already solved the equation. Hence, it’s the only solution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You are claim that $log_a b=log_A B leftrightarrow a=A cap b=B$.
    $endgroup$
    – Ricardo Largaespada
    Jan 1 at 7:25










  • $begingroup$
    If $A = a^n$ and $B = b^n$, then yes, $log_a b = log_A B$. (Of course, assuming $b > 0$.)
    $endgroup$
    – KM101
    Jan 1 at 7:28
















0












$begingroup$

If you want a purely algebraic solution, although rather tedious, you can solve the equation as follows:



$$log_5(2x+1) = frac{log_3(2x+1)}{log_3 5} tag{1}$$



$$frac{log_3(2x+1)}{log_3 5} = log_3(3x-3) tag{2}$$



$$left(3^{log_3(2x+1)}right)^{frac{1}{log_3 5}} = 3x-3 tag{3}$$



$$(2x+1)^{frac{1}{log_3 5}} = 3x-3 tag{4}$$



$$(2x+1)^{log_5 3} = 3x-3 tag{5}$$



$$log_{2x+1} (3x-3) = log_5 3 tag{6}$$



$$boxed{x = 2} tag{7}$$



$(1)$: Change of bases: $log_a b = frac{log_c b}{log_c a}$



$(2)$: Rewriting the equation



$(3)$: Rewriting as an exponential: $c = log_a b iff a^c = b$



$(4)$: Simplifying



$(5)$: Inversion of base and argument: $frac{1}{log_a b} = log_b a$



$(6)$: Rewriting as a logarithm: $a^b = c iff log_a b = c$



$(7)$: Solution to $2x+1 = 5$ and $3x-3 = 3$



But of course, you have to check for any extra solutions. Since $log_a b^c = clog_a b$ and $log_{a^c} b = frac{1}{c}log_a b$, you have $$log_{a^ c} b^c = log_a b$$



$$implies log_5 3 = log_{5^n} 3^n$$



$$implieslog_{2x+1} (3x-3) = log_{5^n} 3^n$$



$$implies 2x+1 = 5^n; quad 3x-3 = 3^n$$



$$3x-3 = 3^n implies x-1 = 3^{n-1} implies x = 3^{n-1}+1$$



Plugging this in the first equation, you have



$$2left(3^{n-1}+1right) = 5^n implies 2left(3^{n-1}right)+3 = 5^n$$



It’s fairly obvious the only solution is $n = 1$ (since the RHS grows at a much faster rate), for which we already solved the equation. Hence, it’s the only solution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You are claim that $log_a b=log_A B leftrightarrow a=A cap b=B$.
    $endgroup$
    – Ricardo Largaespada
    Jan 1 at 7:25










  • $begingroup$
    If $A = a^n$ and $B = b^n$, then yes, $log_a b = log_A B$. (Of course, assuming $b > 0$.)
    $endgroup$
    – KM101
    Jan 1 at 7:28














0












0








0





$begingroup$

If you want a purely algebraic solution, although rather tedious, you can solve the equation as follows:



$$log_5(2x+1) = frac{log_3(2x+1)}{log_3 5} tag{1}$$



$$frac{log_3(2x+1)}{log_3 5} = log_3(3x-3) tag{2}$$



$$left(3^{log_3(2x+1)}right)^{frac{1}{log_3 5}} = 3x-3 tag{3}$$



$$(2x+1)^{frac{1}{log_3 5}} = 3x-3 tag{4}$$



$$(2x+1)^{log_5 3} = 3x-3 tag{5}$$



$$log_{2x+1} (3x-3) = log_5 3 tag{6}$$



$$boxed{x = 2} tag{7}$$



$(1)$: Change of bases: $log_a b = frac{log_c b}{log_c a}$



$(2)$: Rewriting the equation



$(3)$: Rewriting as an exponential: $c = log_a b iff a^c = b$



$(4)$: Simplifying



$(5)$: Inversion of base and argument: $frac{1}{log_a b} = log_b a$



$(6)$: Rewriting as a logarithm: $a^b = c iff log_a b = c$



$(7)$: Solution to $2x+1 = 5$ and $3x-3 = 3$



But of course, you have to check for any extra solutions. Since $log_a b^c = clog_a b$ and $log_{a^c} b = frac{1}{c}log_a b$, you have $$log_{a^ c} b^c = log_a b$$



$$implies log_5 3 = log_{5^n} 3^n$$



$$implieslog_{2x+1} (3x-3) = log_{5^n} 3^n$$



$$implies 2x+1 = 5^n; quad 3x-3 = 3^n$$



$$3x-3 = 3^n implies x-1 = 3^{n-1} implies x = 3^{n-1}+1$$



Plugging this in the first equation, you have



$$2left(3^{n-1}+1right) = 5^n implies 2left(3^{n-1}right)+3 = 5^n$$



It’s fairly obvious the only solution is $n = 1$ (since the RHS grows at a much faster rate), for which we already solved the equation. Hence, it’s the only solution.






share|cite|improve this answer









$endgroup$



If you want a purely algebraic solution, although rather tedious, you can solve the equation as follows:



$$log_5(2x+1) = frac{log_3(2x+1)}{log_3 5} tag{1}$$



$$frac{log_3(2x+1)}{log_3 5} = log_3(3x-3) tag{2}$$



$$left(3^{log_3(2x+1)}right)^{frac{1}{log_3 5}} = 3x-3 tag{3}$$



$$(2x+1)^{frac{1}{log_3 5}} = 3x-3 tag{4}$$



$$(2x+1)^{log_5 3} = 3x-3 tag{5}$$



$$log_{2x+1} (3x-3) = log_5 3 tag{6}$$



$$boxed{x = 2} tag{7}$$



$(1)$: Change of bases: $log_a b = frac{log_c b}{log_c a}$



$(2)$: Rewriting the equation



$(3)$: Rewriting as an exponential: $c = log_a b iff a^c = b$



$(4)$: Simplifying



$(5)$: Inversion of base and argument: $frac{1}{log_a b} = log_b a$



$(6)$: Rewriting as a logarithm: $a^b = c iff log_a b = c$



$(7)$: Solution to $2x+1 = 5$ and $3x-3 = 3$



But of course, you have to check for any extra solutions. Since $log_a b^c = clog_a b$ and $log_{a^c} b = frac{1}{c}log_a b$, you have $$log_{a^ c} b^c = log_a b$$



$$implies log_5 3 = log_{5^n} 3^n$$



$$implieslog_{2x+1} (3x-3) = log_{5^n} 3^n$$



$$implies 2x+1 = 5^n; quad 3x-3 = 3^n$$



$$3x-3 = 3^n implies x-1 = 3^{n-1} implies x = 3^{n-1}+1$$



Plugging this in the first equation, you have



$$2left(3^{n-1}+1right) = 5^n implies 2left(3^{n-1}right)+3 = 5^n$$



It’s fairly obvious the only solution is $n = 1$ (since the RHS grows at a much faster rate), for which we already solved the equation. Hence, it’s the only solution.







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answered Jan 1 at 6:40









KM101KM101

6,0861525




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  • $begingroup$
    You are claim that $log_a b=log_A B leftrightarrow a=A cap b=B$.
    $endgroup$
    – Ricardo Largaespada
    Jan 1 at 7:25










  • $begingroup$
    If $A = a^n$ and $B = b^n$, then yes, $log_a b = log_A B$. (Of course, assuming $b > 0$.)
    $endgroup$
    – KM101
    Jan 1 at 7:28


















  • $begingroup$
    You are claim that $log_a b=log_A B leftrightarrow a=A cap b=B$.
    $endgroup$
    – Ricardo Largaespada
    Jan 1 at 7:25










  • $begingroup$
    If $A = a^n$ and $B = b^n$, then yes, $log_a b = log_A B$. (Of course, assuming $b > 0$.)
    $endgroup$
    – KM101
    Jan 1 at 7:28
















$begingroup$
You are claim that $log_a b=log_A B leftrightarrow a=A cap b=B$.
$endgroup$
– Ricardo Largaespada
Jan 1 at 7:25




$begingroup$
You are claim that $log_a b=log_A B leftrightarrow a=A cap b=B$.
$endgroup$
– Ricardo Largaespada
Jan 1 at 7:25












$begingroup$
If $A = a^n$ and $B = b^n$, then yes, $log_a b = log_A B$. (Of course, assuming $b > 0$.)
$endgroup$
– KM101
Jan 1 at 7:28




$begingroup$
If $A = a^n$ and $B = b^n$, then yes, $log_a b = log_A B$. (Of course, assuming $b > 0$.)
$endgroup$
– KM101
Jan 1 at 7:28


















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