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I need to solve the equation below for x

$$ \log_a x = log_{a^2}(x+6) $$ Note that the base a itself is squared in the RHS of the equation

I've thought about tackling the LHS and RHS separately and end up with a quadratic with two solutions of x, one of which is negative, which can't be right

Costa58
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3 Answers3

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Hint:

$\log_ax=\log_{a^2}(x+6)=k$(say)

$a^k=x,x+6=(a^2)^k=(a^k)^2=x^2$

Remember, for real analysis, we need $x,x+6>0$ assuming $a>0$

lab bhattacharjee
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  • Ok it looks like my method was right, I did get two solutions of -2 and 3, but thought that both had to be valid, instead of rejecting the -2 solution – Costa58 Sep 28 '17 at 13:22
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$$\frac{ln(x)}{\ln(a)}=\frac{\ln(x+6)}{\ln(a^2}$$ for $$a>0$$ and $a\neq 1$ we get $$2\ln(x)=\ln(x+6)$$ threrefore $$x^2=x+6$$ can you solve this?

Dr. Sonnhard Graubner
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  • Yep, I got there by another way, but because I had two solutions, one of which was negative, I thought the whole answer was wrong instead of just rejecting that negative solution – Costa58 Sep 28 '17 at 13:23
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Hint: $\log_{a^2}(x+6)=\frac{\log_{a}(x+6)}{\log_a(a^2)}=\frac{\log_{a}(x+6)}{2}$.

Can you proceed ?

Fred
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