Is $frac{ln2^{n+1}}{1}$ equal to $frac{ln2^{n}*ln2^{1}}{1}$ or $frac{ln2^{n}2^{1}}{1}$












-1












$begingroup$


Well i am not sure about this.
I have something like this $$frac{ln2^{n+1}}{1}$$
And i dont know what is corret can i extend it like this?
$$frac{ln2^{n}*ln2^{1}}{1}$$
Or this one is correct?
$$frac{ln2^{n}2^{1}}{1}$$






algebra-precalculus logarithms







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$endgroup$












  • $begingroup$
    This is not at all clear. Why do you write all those $1's$ as denominators? Of course $ln (2^{n+1})=ln (2^ntimes 2^1)$ since $2^{n+1}=2^ntimes 2^1$. Is that what you are asking?
    $endgroup$
    – lulu
    Dec 19 '18 at 17:44










  • $begingroup$
    both are wrong $ln(acdot b)=ln a + ln b$, $ln 2^{n+1}=ln 2^n+ln 2=(n+1)ln 2$
    $endgroup$
    – Vasya
    Dec 19 '18 at 17:44










  • $begingroup$
    @Vasya I simply assumed $ln 2^n2^1$ to mean $ln (2^n2^1)$. But it isn't clear what the OP meant.
    $endgroup$
    – fleablood
    Dec 19 '18 at 20:07
















-1












$begingroup$


Well i am not sure about this.
I have something like this $$frac{ln2^{n+1}}{1}$$
And i dont know what is corret can i extend it like this?
$$frac{ln2^{n}*ln2^{1}}{1}$$
Or this one is correct?
$$frac{ln2^{n}2^{1}}{1}$$






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is not at all clear. Why do you write all those $1's$ as denominators? Of course $ln (2^{n+1})=ln (2^ntimes 2^1)$ since $2^{n+1}=2^ntimes 2^1$. Is that what you are asking?
    $endgroup$
    – lulu
    Dec 19 '18 at 17:44










  • $begingroup$
    both are wrong $ln(acdot b)=ln a + ln b$, $ln 2^{n+1}=ln 2^n+ln 2=(n+1)ln 2$
    $endgroup$
    – Vasya
    Dec 19 '18 at 17:44










  • $begingroup$
    @Vasya I simply assumed $ln 2^n2^1$ to mean $ln (2^n2^1)$. But it isn't clear what the OP meant.
    $endgroup$
    – fleablood
    Dec 19 '18 at 20:07














-1












-1








-1





$begingroup$


Well i am not sure about this.
I have something like this $$frac{ln2^{n+1}}{1}$$
And i dont know what is corret can i extend it like this?
$$frac{ln2^{n}*ln2^{1}}{1}$$
Or this one is correct?
$$frac{ln2^{n}2^{1}}{1}$$






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$




Well i am not sure about this.
I have something like this $$frac{ln2^{n+1}}{1}$$
And i dont know what is corret can i extend it like this?
$$frac{ln2^{n}*ln2^{1}}{1}$$
Or this one is correct?
$$frac{ln2^{n}2^{1}}{1}$$






algebra-precalculus logarithms




algebra-precalculus logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 18:51









Larry

2,53031131




2,53031131










asked Dec 19 '18 at 17:40









Juraj JakubovJuraj Jakubov

187




187












  • $begingroup$
    This is not at all clear. Why do you write all those $1's$ as denominators? Of course $ln (2^{n+1})=ln (2^ntimes 2^1)$ since $2^{n+1}=2^ntimes 2^1$. Is that what you are asking?
    $endgroup$
    – lulu
    Dec 19 '18 at 17:44










  • $begingroup$
    both are wrong $ln(acdot b)=ln a + ln b$, $ln 2^{n+1}=ln 2^n+ln 2=(n+1)ln 2$
    $endgroup$
    – Vasya
    Dec 19 '18 at 17:44










  • $begingroup$
    @Vasya I simply assumed $ln 2^n2^1$ to mean $ln (2^n2^1)$. But it isn't clear what the OP meant.
    $endgroup$
    – fleablood
    Dec 19 '18 at 20:07


















  • $begingroup$
    This is not at all clear. Why do you write all those $1's$ as denominators? Of course $ln (2^{n+1})=ln (2^ntimes 2^1)$ since $2^{n+1}=2^ntimes 2^1$. Is that what you are asking?
    $endgroup$
    – lulu
    Dec 19 '18 at 17:44










  • $begingroup$
    both are wrong $ln(acdot b)=ln a + ln b$, $ln 2^{n+1}=ln 2^n+ln 2=(n+1)ln 2$
    $endgroup$
    – Vasya
    Dec 19 '18 at 17:44










  • $begingroup$
    @Vasya I simply assumed $ln 2^n2^1$ to mean $ln (2^n2^1)$. But it isn't clear what the OP meant.
    $endgroup$
    – fleablood
    Dec 19 '18 at 20:07
















$begingroup$
This is not at all clear. Why do you write all those $1's$ as denominators? Of course $ln (2^{n+1})=ln (2^ntimes 2^1)$ since $2^{n+1}=2^ntimes 2^1$. Is that what you are asking?
$endgroup$
– lulu
Dec 19 '18 at 17:44




$begingroup$
This is not at all clear. Why do you write all those $1's$ as denominators? Of course $ln (2^{n+1})=ln (2^ntimes 2^1)$ since $2^{n+1}=2^ntimes 2^1$. Is that what you are asking?
$endgroup$
– lulu
Dec 19 '18 at 17:44












$begingroup$
both are wrong $ln(acdot b)=ln a + ln b$, $ln 2^{n+1}=ln 2^n+ln 2=(n+1)ln 2$
$endgroup$
– Vasya
Dec 19 '18 at 17:44




$begingroup$
both are wrong $ln(acdot b)=ln a + ln b$, $ln 2^{n+1}=ln 2^n+ln 2=(n+1)ln 2$
$endgroup$
– Vasya
Dec 19 '18 at 17:44












$begingroup$
@Vasya I simply assumed $ln 2^n2^1$ to mean $ln (2^n2^1)$. But it isn't clear what the OP meant.
$endgroup$
– fleablood
Dec 19 '18 at 20:07




$begingroup$
@Vasya I simply assumed $ln 2^n2^1$ to mean $ln (2^n2^1)$. But it isn't clear what the OP meant.
$endgroup$
– fleablood
Dec 19 '18 at 20:07










1 Answer
1






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oldest

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1












$begingroup$

Note that for $a,b>0$, we have that
$$
ln(ab)=ln(a)+ln(b).
$$
In particular
$$
ln(2^{n+1})=ln(2^ntimes 2)=ln(2^n)+ln 2
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ou thank you very much :)
    $endgroup$
    – Juraj Jakubov
    Dec 19 '18 at 17:51












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Note that for $a,b>0$, we have that
$$
ln(ab)=ln(a)+ln(b).
$$
In particular
$$
ln(2^{n+1})=ln(2^ntimes 2)=ln(2^n)+ln 2
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ou thank you very much :)
    $endgroup$
    – Juraj Jakubov
    Dec 19 '18 at 17:51
















1












$begingroup$

Note that for $a,b>0$, we have that
$$
ln(ab)=ln(a)+ln(b).
$$
In particular
$$
ln(2^{n+1})=ln(2^ntimes 2)=ln(2^n)+ln 2
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ou thank you very much :)
    $endgroup$
    – Juraj Jakubov
    Dec 19 '18 at 17:51














1












1








1





$begingroup$

Note that for $a,b>0$, we have that
$$
ln(ab)=ln(a)+ln(b).
$$
In particular
$$
ln(2^{n+1})=ln(2^ntimes 2)=ln(2^n)+ln 2
$$






share|cite|improve this answer









$endgroup$



Note that for $a,b>0$, we have that
$$
ln(ab)=ln(a)+ln(b).
$$
In particular
$$
ln(2^{n+1})=ln(2^ntimes 2)=ln(2^n)+ln 2
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 17:44









Foobaz JohnFoobaz John

22.9k41552




22.9k41552












  • $begingroup$
    Ou thank you very much :)
    $endgroup$
    – Juraj Jakubov
    Dec 19 '18 at 17:51


















  • $begingroup$
    Ou thank you very much :)
    $endgroup$
    – Juraj Jakubov
    Dec 19 '18 at 17:51
















$begingroup$
Ou thank you very much :)
$endgroup$
– Juraj Jakubov
Dec 19 '18 at 17:51




$begingroup$
Ou thank you very much :)
$endgroup$
– Juraj Jakubov
Dec 19 '18 at 17:51


















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