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There is a Equation: $2^x + 3^x = 6^x + 6$. I tried to do a lot of thing but I couldn't solve this equation. one of the thing that I do:
I assume that
$t_1 = 2^x$ so $\log_2 t_1 = x$
$t_2 = 3^x$ so $\log_3 t_2 = x$
$t_3 = 6^x$ so $\log_6 t_3 = x$
so we have
$t_1 + t_2 = t_3 + 6$ and
$\log_2 t_1 = \log_3 t_2 = \log_6 t_3 = \log_6 t_1*t_2$
Then I try lots of way to solve my equation with this method but nothing happen!

Is it possible to help me to solve this? I'm sorry for bad English too.
Thanks.

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    $\begingroup$ this equation has no real roots $\endgroup$ – Dr. Sonnhard Graubner Jan 7 '17 at 19:02
  • $\begingroup$ x is a real or an integer ? You should start by studying the function $f(x)=6^x-3^x -2^x-6$ and search for its zeros $\endgroup$ – allced Jan 7 '17 at 19:04
  • $\begingroup$ Yes. why? I know that $6 + 6^x$ is even and $2^x + 3^x$ is odd so it doesn't have any answer in Integer but why about real number? How to prove that? $\endgroup$ – Amin Jan 7 '17 at 19:07
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Write the equation as $(2^x-1)(3^x-1) = -5$. If $x \geq 0$ then both factors on the left hand side are nonnegative, and if $x<0$ then both factors are negative. Either way, the product is nonnegative, hence it can't be $-5$.

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  • $\begingroup$ I think that it has answer in real number but it doesn't have. Thanks for answer. $\endgroup$ – Amin Jan 7 '17 at 19:10

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