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Given is the equation:

$$\log_x3+\log_x12 = 2$$

How do I solve it? My idea was to use the formula $\log_a(b) = \frac{\ln b}{\ln a}$ but that does not seem to help here.

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Your idea is fine. Keep running with it. – Qiaochu Yuan Apr 24 '13 at 1:25

Hint: $\log_x3 + \log_x12 = \log_x (3 \times 12) = \log_x 36$.

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Your method is feasible indeed! Using the formula $\log_ab=\frac{\ln a}{\ln b}$, we get $$\frac{\ln 3}{\ln x}+\frac{\ln 12}{\ln x}=2 \\ \frac{\ln 3+\ln 12}{\ln x}=2 \\ \frac{\ln(3\times 12)}{\ln x}=2 \\ \ln x=\frac 12\times \ln36 \\ \ln x=\ln{36^\frac 12} \\ \ln x = \ln 6 \\ x=6$$ Of course, change of base to natural logarithm is not a must in this question as in the answer given by @Ross B. But I would like to tell you that your approach is definitely fine.

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Maybe this way is easier to understand: $$\log_x3+\log_x12=2$$ $$\therefore\quad x^{\log_x3+\log_x12}=x^2$$ $$\therefore\quad x^{\log_x3}\cdot x^{\log_x12}=x^2$$ $$\therefore\quad 3\cdot12=x^2$$ $$\therefore\quad x^2=36=6^2$$ Since $x$ must be positive (since $\log_x$ must be defined), you get $x=6$.

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my answer,


$\log_x(3\times 12)=2\log_xx$


compare the numbers on both the sides,

$36=x^2\ \ \ \ $ forall $(0<x<1$ or $x>1)$


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Isn't your answer similar to @RossB.'s? – Mohsen Shahriari Dec 25 '15 at 8:52
Oops! I didn't see that before posting my answer. should i delete it? @MohsenShahriari – Bhaskara-III Dec 25 '15 at 17:53
That's up to you! – Mohsen Shahriari Dec 25 '15 at 18:12

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