Let $4\cdot7^{x+2}=9^{2x-3}.$
I do not know how to solve for $x$.
Progress
Took logarithms, got $$4(x+2\log7)=(2x-3)\log9$$ $$(x+2)\log7=[(2x-3)\log9]/4$$
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$\begingroup$ what did you try to do? $\endgroup$– Ofir SchnabelJan 13, 2015 at 8:39
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$\begingroup$ Do you mean $4(7^x + 2) = 9^{2x}-3$ or $4(7^{x+2}) = 9^{2x-3}$? $\endgroup$– EmpiricistJan 13, 2015 at 8:40
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$\begingroup$ 4(x+2log7)=(2x-3)log9 and (x+2)log7=[(2x-3)log9]/4 $\endgroup$– RuiJan 13, 2015 at 8:42
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$\begingroup$ @SRX The second one. Sorry! $\endgroup$– RuiJan 13, 2015 at 8:43
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$\begingroup$ I've edited your question @Rui. $\endgroup$– Alex SilvaJan 13, 2015 at 8:43
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2 Answers
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It is clarified that we are to solve $4(7^{x+2}) = 9^{2x-3}$. Taking $\log$ is the right process but you have to make sure you do it right.
Using $\log(ab) = \log(a) + \log(b)$, the L.H.S. should be $\log 4 + (x+2) \log 7$ but not $4(x + 2) \log 7$.
Hence we have
\begin{align} \log 4 + (x+2) \log 7 & = (2x - 3) \log 9 \\ (\log 7)x + (\log 4 + 2\log 7) & = (2\log 9) x - 3 \log 9 \\ (2\log 9- \log 7)x & = \log 4 + 2\log 7 + 3 \log 9 \\ x & = \frac{\log 4 + 2\log 7 + 3 \log 9}{2\log 9- \log 7} \end{align}
answered Jan 13, 2015 at 8:46
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$\begingroup$ Your RHS isn't quite right, it should be log(4) + (x + 2)log(7) $\endgroup$– PM 2RingJan 13, 2015 at 8:49
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$\begingroup$ I start from the question provided in the comment, and the equations are inconsistent with the one originally posed. $\endgroup$ Jan 13, 2015 at 8:54
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$$4 \times {7^{x + 2}} = {9^{2x - 3}} \Leftrightarrow 4 \times 49 \times {7^x} = \frac{{{{81}^x}}} {{729}} \Leftrightarrow {\left( {\frac{{81}} {7}} \right)^x} = 4 \times 49 \times 729 = 142884 \Leftrightarrow x = {\log _{81/7}}142884.$$
