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Please help me to solve for x using maybe logarithm or exponential rules (or both)

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

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closed as off-topic by Andres Caicedo, Weapon of Choice, Fly by Night, PVAL, anorton Aug 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

3 Answers 3

up vote 2 down vote accepted

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

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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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