Rewriting $(\log_{11}5)/(\log_{11} 15)$

Question

I'm struggling with this can anyone tell me the solution of this?

$$ \frac{\log_{11}{5}}{\log_{11}{15}} $$

A) $\log_{11}{15}$

B) $\log_{11}{5}$

C) $\log_{5}{15}$

D) $\log_{15}{5}$

$ log _b a = \dfrac{log_u a} {log_u b} $ where $u$ is any real number. Read that backwards. — Narasimham, Jan 14 '18 at 16:51
As a multiple choice Question, one way to add context is to identify one or more of the answers posed that you are able to eliminate. Are there some? — hardmath, Jan 14 '18 at 17:38

score 0 · Answer 1 · answered Jan 14 '18 at 15:22

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Hint. You have $$ \log_{11}5=\frac{\log 5}{\log 11},\quad \log_{11}{15}=\frac{\log 15}{\log 11}. $$

answered Jan 14 '18 at 15:22

Olivier Oloa

120,989

score 0 · Answer 2 · answered Jan 14 '18 at 15:22

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note that $$\log_{15} 5=\frac{\log_a 5}{\log_a 15}$$ where $a>0$ and $a\ne 1$

answered Jan 14 '18 at 15:22

Dr. Sonnhard Graubner

95,283

score 0 · Accepted Answer · answered Jan 14 '18 at 15:59

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$$\frac{Log_ca}{Log_cb}=Log_ba$$ $$\frac{Log_{11}5}{Log_{11}{15}}=Log_{15}{5}$$

answered Jan 14 '18 at 15:59

Mostafa Ayaz

31,924

Are u sure? I think answer is C.? – Muhammad Ahmed Jan 14 '18 at 16:01
Which c? :) ......... – Mostafa Ayaz Jan 14 '18 at 16:03
are u sure that $Log_5 15$ is correct? – Muhammad Ahmed Jan 14 '18 at 16:04
Yes! Also the answer must intuitively be less than one... – Mostafa Ayaz Jan 14 '18 at 16:05

Michael Rybkin · Answer 4 · 2018-01-14T16:34:58.337

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Have you ever seen the change of base formula for logarithms before: $\log_{a}{b}=\frac{\log_{x}{b}}{\log_{x}{a}}$? Watch this video to find out what it is and how to use it.

$$ \frac{\log_{11}{5}}{\log_{11}{15}}=\log_{15}{5} $$

So, the answer is D.

edited Jan 14 '18 at 16:34

answered Jan 14 '18 at 16:28

Michael Rybkin

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Rewriting $(\log_{11}5)/(\log_{11} 15)$

4 Answers4