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If $\;\log_{12}(a^2+b)=2+\log_3b=3+\log_2a$, what is the value of $\sqrt{\frac1b+\frac1{a^2}}$ ?

I used change of base,

$$\log_{12}(a^2+b)=\log_3(9b)\;\Rightarrow\; \frac{\log_3(a^2+b)}{1+2\log_32}=\log_3(9b)$$ $$\log_{12}(a^2+b)=\log_2(8a)\;\Rightarrow\; \frac{\log_2(a^2+b)}{2+\log_23}=\log_2(8a)$$ $$\log_3b=\log_2(2a) $$ I managed to get same bases in numerator of first two equations. But not sure if it helps since it is getting uglier.

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1 Answer 1

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let $$\log_{12}{(a^2+b)}=2+\log_{3}{b}=3+\log_{2}{a}=t$$ then we have $$a^2+b=12^t,9b=3^t,8a=2^t$$ so $$\sqrt{\dfrac{1}{b}+\dfrac{1}{a^2}}=\sqrt{\dfrac{a^2+b}{a^2b}}=\sqrt{\dfrac{12^t}{4^{t-3}3^{t-2}}}=24$$

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