Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the right solution for $\log_7x+\log_{\frac17}x^2=\log_{49}x-3$. What logarithm identities used?

share|cite|improve this question
I edited it to make it more readable. I don't think I missed anything, but if you think I did you can change it back. – Amr Dec 9 '12 at 16:37
@Akos, is the base of log in the right hand side $49?$ – lab bhattacharjee Dec 9 '12 at 16:45
The base is 49 on the right side. – Ákos Kovács Dec 9 '12 at 16:47
I edited the question, now everything well formatted. – Ákos Kovács Dec 9 '12 at 16:53
up vote 1 down vote accepted

You will need the following identities: (For all $a,b,c>0$, $n$ real) $$\begin{array}{l}(1) \hspace{10pt }\log_ab=x\Leftrightarrow a^x=b\\ (2) \hspace{10pt }\log_ab=\frac{\log_cb}{\log_ca}\\ (3) \hspace{10pt } n\log_ab=\log_a(b^n)\end{array}$$ Hence $\log_{\frac17}x^2=\frac{\log x^2}{\log \frac17}=\frac{\log x^2}{-\log 7}=-\log_7x^2=-2\log_7x$
and $\log_{49} x=\log_{7^2} x=\frac12\log_7x$
So $\log_7x+\log_{\frac17}x^2=-\log_7x$. Hence you have: $$-\log_7x=\frac12\log_7x - 3 \hspace{5pt}\Rightarrow\hspace{5pt} \frac32\log_7x=3\hspace{5pt}\Rightarrow\hspace{5pt}\log_7x=2$$

share|cite|improve this answer

Apply $\log_ab=\frac{\log b}{\log a}$ and $\log b^m=m\log b$


becomes, $$\frac{\log x}{\log 7}-2\frac{\log x}{\log 7}=\frac{\log x}{2\log 7}-3$$

So, $\log_7x-2\log_7x=\frac12\log_7x-3,\implies \log_7x=2, x=7^2=49$

share|cite|improve this answer

$$\log_{a}b=\dfrac{\log_c b}{ \log_c a}$$ $$log_7 x + log_{1/7} x^2 =log_49 x - 3 $$

So, $$\log_{1/7}x^2=\dfrac{\log_7 x^2}{ \log_7 1/7}$$ $$\log_{1/7}x^2={-\log_7 x^2}$$

share|cite|improve this answer

Even if you don't know properties of logs, you can procced as follows (this is the idea behind those proofs):

$$7^{\log_7 x + \log_{1/7} x^2} =7^{\log_{49} x - 3}$$

$$7^{\log_7 x} =x$$ $$7^{ \log_{1/7} x^2} = [\left( \frac{1}{7}\right)^{\log_{1/7} x^2}]^{-1}=x^{-2}$$ $$7^{\log_49 x - 3} = \sqrt{49^{\log_{49} x - 3}}=\sqrt{x-3}$$

share|cite|improve this answer

$$\log_7 x + \log_{1/7} x^2 =\log_{7^2} (x) - 3 $$

$$\log_7 x + \frac{\log_7(1/7)}{\log_7 x^2} =\frac{\log_77^2}{\log_7 (x) }-3 $$

$$\log_7 x + \frac{-1}{\log_7 x^2} =\frac{2}{\log_7 (x) }-3 $$

$\log_7 x=t$ then we have


$$t^2+3t-3=0$$ $$\log_7x_1=\frac{-3+\sqrt{21}}{2},x_1=7^{\frac{-3+\sqrt{21}}{2}}$$ $$\log_7x_2=\frac{-3-\sqrt{21}}{2},x_2=7^{\frac{-3-\sqrt{21}}{2}}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.