<s>[INST] Solve $3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$ Can someone point me in the right direction how to solve this?
$3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$
I guess I have to get to logarithms of the same base. But how? What principle should I use here?
Thx
 [/INST] $$3^{ x+2 }\cdot 4^{ -(x+3) }+3^{ x+4 }\cdot 4^{ -(x+3) }=\frac { 40 }{ 9 } \\ { 4 }^{ -\left( x+3 \right)  }{ 3 }^{ x+2 }\left( 1+9 \right) =\frac { 40 }{ 9 } \\ { 4 }^{ -\left( x+3 \right)  }{ 3 }^{ x+2 }=\frac { 4 }{ 9 } \\ { \left( \frac { 3 }{ 4 }  \right)  }^{ x }\frac { 9 }{ 64 } =\frac { 4 }{ 9 } \\ { \left( \frac { 3 }{ 4 }  \right)  }^{ x }={ \left( \frac { 4 }{ 3 }  \right)  }^{ 4 }\\ x=-4\\ \\ $$
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