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Was solving using properties of logarithm but got stuck at the equation $x\log 3=\log5+\log5$

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    $\begingroup$ So you have $x(a)=2b$ and you have to solve for $x$. $\endgroup$
    – Anurag A
    Jul 14, 2020 at 14:12
  • $\begingroup$ Yes dont know the answer $\endgroup$
    – Maths
    Jul 14, 2020 at 14:14
  • $\begingroup$ If that's a minus sign on left need to adjust solution. $\endgroup$
    – coffeemath
    Jul 14, 2020 at 14:15
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    $\begingroup$ Ask yourself,how would you solve for $x$ in $5x=7$? $\endgroup$
    – Anurag A
    Jul 14, 2020 at 14:15
  • $\begingroup$ We can simply write $x=\log_3(5^2)$ :) $\endgroup$
    – K.defaoite
    Jul 14, 2020 at 14:21

4 Answers 4

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Hint: $\log 3$ and $\log 5$ in your equation are just constants. How would you solve for $x$ if they were replaced by say, $1$ and $2$?

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  • $\begingroup$ Thanks a lot paulino for helping $\endgroup$
    – Maths
    Jul 14, 2020 at 14:21
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As a side note to the previous answers, $x = \frac{2\log 5}{\log 3}$ can be further simplified in the denominator to $x = \frac{\log 25}{\log 3}$, and using log rules $x = \log_3{25}$. This is a way to remove the fraction.

-FruDe

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  • $\begingroup$ Thank u for sol $\endgroup$
    – Maths
    Jul 14, 2020 at 18:09
  • $\begingroup$ Sure thing @Maths $\endgroup$ Jul 14, 2020 at 18:16
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you are almost done $$x\log3=\log5+\log5=2\log5$$ $$x=\frac{2\log5}{\log3}$$ $$x=\frac{2\cdot 0.6989}{0.4771}\approx2.93$$

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    $\begingroup$ Thanks for helping $\endgroup$
    – Maths
    Jul 14, 2020 at 14:22
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$$3^x=5^2\implies \log3^x=\log5^2\implies x\log3=2\log5\implies x=\frac{2\log5}{\log3}$$ I hope that helps :)

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