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I'm trying to answer this problem:

Write $$2+\log_3 x+\log_9 x^4-\log_{27} x^5$$

into one single logarithm.

This is what I'm stuck with:

$$\frac{\log x}{\log 3}+ \frac{\log x^4}{\log 3^2}+\frac{\log x^5}{\log 3^3}$$

I don't know how to advance from this and I don't know what to do with the 2

Possible answers: A) $\log 1$ B) $\log2x^5$ C) $\log_3 9x^\frac{4}{3}$ D) $\log_3 9x^\frac{10}{3}$

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  • $\begingroup$ Why is the question in the title so different from the one in the question? $\endgroup$ – Kabo Murphy Aug 3 at 12:39
  • $\begingroup$ @KaviRamaMurthy They're the same, at least now. $\endgroup$ – Parcly Taxel Aug 3 at 12:43
  • $\begingroup$ Use $\log x$ (written as \$\log x\$) to help make the expressions neater and the spacing better. The MathJax reference can be found here. $\endgroup$ – Toby Mak Aug 3 at 12:55
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We have $$\log_9x=\frac{\ln x}{\ln 9}=\frac{\ln x}{2\ln 3}=\frac12\log_3x$$ and similarly $\log_{27}x=\frac13\log_3x$. Thus $$2+\log_3x+\log_9x^4-\log_{27}x^5=2+\log_3x+\frac12\log_3x^4-\frac13\log_3x^5$$ $$=2+\log_3x+2\log_3x-\frac53\log_3x=\log_39+\frac43\log_3x=\log_39x^{4/3}$$ and the answer is (c).

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Trick answer

Substitute $x=3$ into the expression to get:

$$2 + \log_3 3 + \log_{3^2} 3^4 - \log_{3^3} 3^5$$ $$=2+1+2-\frac{5}{3} = \frac{10}{3}$$

Now substitute $x=3$ into the multiple choice options and see which option matches.

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Except the sign, you did it correctly, and $2=\log_3{3^2},$ thus $$ \begin{aligned} 2+\log_3 x+\log_9 x^4-\log_{27} x^5&=\frac{\log 3^2}{\log 3}+\frac{\log x}{\log 3}+ \frac{4\log x}{2\log 3}-\frac{5\log x}{3\log 3}\\&=\frac{\log 3^2}{\log 3}+\frac{\log x}{6\log 3}\left( 6+12-10\right)\\&=\frac{\log 3^2}{\log 3}+\frac{4\log x}{3\log 3}\\&=\frac{\log 9 +\log x^{4\over 3}}{\log 3}\end{aligned}$$ I am sure, you can finish it.

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The answer is $\log_3(3^{2}x x^{4/2}x^{-5/3})$. Can you figure out why this is true? The correct answer is C).

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Since $x>0$ we obtain: $$2+\log_3 x+\log_9 x^4-\log_{27} x^5=2+\log_3 x+\frac{4}{2}\log_3 x-\frac{5}{3}\log_{3} x=$$ $$=\log_39+\frac{4}{3}\log_3x=\log_39x^{\frac{4}{3}}.$$

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