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So I was again looking through the homepage of Youtube to see if there were any math video that I thought that I might be able to solve when I came across this video by Learncommunolizer asking for the value of $x$ in $$\color{black}{9^x=18}$$which I thought I might be able solve. Here is my attempt at solving the aforementioned question:$$9^x=18$$$$3^{2x}=18$$$$3^{2x-1}=6$$$$3^{2x-2}=2$$$$(2x-2)\ln3=\ln2$$$$2x-2=\dfrac{\ln2}{\ln3}$$$$2x=\dfrac{\ln2}{\ln3}+2$$And knowing that$$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{cd}$$Also$$2=\dfrac{2}{1}$$$$2x=\dfrac{\ln(2)+2\ln(3)}{\ln(3)}$$$$\therefore x=\dfrac{\ln(2)+2\ln(3)}{2\ln(3)}$$My question


Is the solution that I have achieved correct, or what could I do to achieve the correct solution/attain it more quickly?

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    $\begingroup$ shouldnt it be $\frac{ad+bc}{bd}$ $\endgroup$ Commented May 15, 2023 at 14:51
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    $\begingroup$ "And knowing that$$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{cd}$$ This is not correct. Try an example. $\endgroup$ Commented May 15, 2023 at 14:56
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    $\begingroup$ @CrSb0001 User0 is correct. Moreover, you used the formula $\frac{ad + bc}{cd}$ in your calculations with $a = \ln 2$, $b = \ln 3$, $c = 2$, and $d = 1$. Notice that $\frac{ad + bc}{bd} = \frac{\ln 2 + 2\ln 3}{ln 3}$. $\endgroup$ Commented May 15, 2023 at 14:56
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    $\begingroup$ well if that were the case, then 2x=$\frac{ln2+2ln3}{2ln3}$ and x would then be $\frac{ln2+2ln3}{4ln3}$ $\endgroup$ Commented May 15, 2023 at 14:56
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    $\begingroup$ Also your solution is correct with an error of 0.0000000006 $\endgroup$ Commented May 15, 2023 at 14:57

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You can directly compute

$$ x \ln (9) = \ln(18) $$ $$\Longrightarrow x = \frac{\ln (18)}{\ln (9)}$$

Factoring and using logarithm rules we get

$$x = \frac{\ln (2 \cdot 3^2)}{\ln (3^2)}$$ $$\Longrightarrow x = \frac{\ln (2) + 2\ln(3)}{2\ln (3)}$$

So your result is correct, but with a few extra steps in between.

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