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I'm having trouble solving this equation step by step:

$$3^{2x+1} = 3^x + 24$$

I've tried to take the log of both sides but then I am stuck with the right hand side being $\log(3^x + 24)$. I've found the answer to '$x$' by trial and error but cannot arrive at the answer otherwise.

Can anyone please show how to work it out properly?

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    $\begingroup$ Hint: assume $3^x=t$ $\endgroup$ – igumnov Jul 14 '13 at 23:01
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Let $u = 3^x$, then we have:

$$ 3u^2 = u + 24 $$ $$ 3u^2 - u - 24 = 0$$

Which is a quadratic in $u$. Solve, and then use that $u = 3^x$ in order to find $x$.

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  • $\begingroup$ Thanks but how do you get the '3' in the 3u^2 from the original 3^(2x+1)? $\endgroup$ – Gez Bishop Jul 14 '13 at 23:23
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    $\begingroup$ @GezBishop $3^{2x+1} = 3^{2x}\cdot3= 3\cdot(3^x)^2$ $\endgroup$ – Alraxite Jul 14 '13 at 23:30
  • $\begingroup$ Thanks very much everyone! $\endgroup$ – Gez Bishop Jul 14 '13 at 23:31
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You can write $3^{2x+1}-3^x=24$ and factor to get $3^x(3^{x+1}-1)=2^3\cdot 3$. The first factor on the left is the only one that can have a factor $3$, so $x=1$

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    $\begingroup$ That assumes looking for integers solutions... $\endgroup$ – Thomas Andrews Jul 14 '13 at 23:04
  • $\begingroup$ @ThomasAndrews: That is correct. They are often easier to find, and so I often start there. Here it works. $\endgroup$ – Ross Millikan Jul 15 '13 at 0:14

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