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is the transformation of this equation: $$9^x + 6^x = 2× 4^x$$ into this: $$\log_2 (9^x) + \log_2 (6^x)=\log_2 (2×4^x)$$ correct? I want to know because I really want to solve this equation.

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    $\begingroup$ I am afraid not, problem is the left hand side. Logs do not distribute that way $\endgroup$ – imranfat Feb 13 '18 at 17:15
  • $\begingroup$ It is not correct. $$ \log_2 (9^x) + \log_2(6^x) = \log_x(9^x \cdot 6^x). $$ So $\log_2 (9^x) + \log_2(6^x) = \log_x(2\times 4^x)$ is equivalent to $ 9^x\times6^x = 2\times 4^x. $ $${}$$ $\endgroup$ – Michael Hardy Feb 13 '18 at 17:15
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    $\begingroup$ So should it be this instead? $$\log_2 (9^x + 6^x)=\log_2 (2×4^x)$$ $\endgroup$ – Robert874 Feb 13 '18 at 17:17
  • $\begingroup$ Robert874: Please explain how the "answer" you saw fit to accept is addressing the question "is the transformation of this equation: $9^x + 6^x = 2× 4^x$ into this: $\log_2 (9^x) + \log_2 (6^x)=\log_2 (2×4^x)$ correct? I want to know because I really want to solve this equation." $\endgroup$ – Did Feb 13 '18 at 23:03
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HINT: write your equation in the form $$\left(\frac{3}{2}\right)^{2x}+\left(\frac{3}{2}\right)^x=2$$

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    $\begingroup$ (-1). This hint absolutely does not address the question asked, which is whether the asker applied the logarithm correctly. A good answer would correct the error and point out the mistake. $\endgroup$ – user296602 Feb 13 '18 at 21:23
  • $\begingroup$ i will note this in the future! $\endgroup$ – Dr. Sonnhard Graubner Feb 15 '18 at 22:46
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No. One property of logarithms is that $\log(a) + \log(b) =\log(ab).$

So $\log(a + b)$ may not be reduced as you have done.

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It's $f(x)=0$, where $$f(x)=\left(\frac{3}{2}\right)^{2x}+\left(\frac{3}{2}\right)^{x}-2.$$ We see that $f$ increases, which says that our equation has one root maximum.

But, $0$ is a root and we are done!

Your reasoning is wrong because $\log(a+b)$ is not always equal to $\log{a}+\log{b}.$

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