Question

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2^x+3+2^x+1-2^x=2^x-1+68​

Answer 1

A solução é 3.

$\begin{array}{c}\tt 2^{x+3}+2^{x+1}-2^{x}=2^{x-1}+68\\\\ \tt 2^{x}\cdot 2^{3}+2^{x}\cdot 2^{1}-2^{x}=2^{x}\div2^{1}+68\\\\ \tt atribuindo\: valor\: para \: 2^{x}\\\\ \tt\to\tt 2^{x}=k\\\\ \tt \: k\cdot 2^{3}+k\cdot 2^{1}-k=\dfrac{k}{2^{1}}+68\\\\ \tt \: k\cdot 8+k\cdot 2-k=\dfrac{k}{2}+68\\\\ \tt\\\\ \tt 8k+2k-k=\dfrac{k}{2}+68\\\\ \tt mmc\: de \:dois\to 2\\\\ \tt\dfrac{16k}{2}+\dfrac{4k}{2}-\dfrac{2k}{2}=\dfrac{k}{2}+\dfrac{136}{2}\\\\ \tt 16k+4k-2k=k+136\\\\ \tt 16k+4k-2k-k=136\\\\ \tt 17k=136\\\\ \tt k=\dfrac{136}{17}\\\\ \tt k=8\\\\ \tt voltando...\\\\ \tt 2^{x}=k\\\\ \tt 2^{x}=8\\\\ \tt \cancel{2}^x=\cancel{2}^3\\\\ \tt\boxed{\tt x=3}\\\\ \tt S=\{3\}\end{array}$

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Answer 2

$2^{x+3}+2^{x+1}-2^x=2^{x-1}+68\implies$

$\implies2^{x+3}+2^{x+1}-2^x-2^{x-1}=68.$

$\text{ 2^{x-1} em evid \rm\hat{e} ncia:}$

$\implies2^{x-1+4}+2^{x-1+2}-2^{x-1+1}-2^{x-1}=68\implies$

$\implies2^{x-1}(2^4+2^2-2^1-1)=68\implies$

$\implies2^{x-1}(16+4-2-1)=68\implies$

$\implies2^{x-1}(17)=68\implies$

$\implies2^{x-1}=\frac{1}{17}\cdot68\implies$

$\implies2^{x-1}=4\implies$

$\implies2^{x-1}=2^2\implies$

$\implies x-1=2\implies$

$\implies x=2+1\implies\underline{\boxed{x=3}}\,.$