# Solve for $x$ in the equation [closed]

$$5^x=2 \cdot 3^x$$

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## closed as off-topic by Andres Caicedo, Weapon of Choice, Fly by Night, PVAL, anortonAug 6 at 0:13

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Hint: Your equation is equivalent to $\dfrac{5^x}{3^x}=2$. –  Zircht Aug 5 at 20:00

$$5^x=2 \cdot 3^x \Rightarrow \left (\frac{5}{3} \right )^x=2 \Rightarrow \log_{\frac{5}{3}} \left (\frac{5}{3} \right )^x= \log_{\frac{5}{3}} 2 \Rightarrow x= \log_{\frac{5}{3}} 2$$

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I think other answers are overcomplicated: just take logarithms to your favourite base to get $$x\log 5=\log 2+x\log 3$$This is a linear equation for $x$, which you should be able to solve. That is equivalent to what others have said, but the first step is easier.

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The equation can be re-written using the rule that $a^b ={\rm e}^{b \ln(a)} \} \ln(a^b) = b \ln(a)$

$$5^x = 2^1 \cdot 3^x$$

which is expanded to:

$$\ln \left( 5^x \right) = \ln \left( 2 \cdot 3^x \right) = \ln(2) + \ln \left( 3^x \right)$$ $$x \ln(5) = \ln(2) + x \ln(3)$$

and solved for

$$x = \frac{\ln(2)}{\ln\left(\frac{5}{3}\right)}$$

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