How to find the x in this exponential function? Thank you!

Question

How to find the x in this exponential function? Thank you!

$2 (7^{x} + 2^{x - 1}) = 7 (2^{x} - 7^{x - 1})$ $2(7^x+2^(x-1))=7(2^x-7^(x-1))$

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Answer 1 · 2018-05-15T23:32:26

${log}_{3.5} (2)$ $log_3.5(2)$

Explanation:

Distributing the variables on each side gives us
$2 \cdot 7^{x} + 2 \cdot 2^{x - 1} = 7 \cdot 2^{x} - 7 \cdot 7^{x - 1}$ $2*7^x+2*2^(x-1) = 7*2^x-7*7^(x-1)$

We know $a^{x - 1} \cdot a = a^{x}$ $a^(x-1)*a=a^x$ using basic exponent rules, so applying that rule gives us

$2 \cdot 7^{x} + 2^{x} = 7 \cdot 2^{x} - 7^{x}$ $2*7^x+2^x=7*2^x-7^x$
Combine like terms
$3 \cdot 7^{x} = 6 \cdot 2^{x}$ $3*7^x=6*2^x$
$\frac{7^{x}}{2^{x}} = 2$ $7^x/2^x=2$

Using another exponent rule, we know $\frac{a^{x}}{b^{2}} = {(\frac{a}{b})}^{x}$ $a^x/b^2=(a/b)^x$ , so
${(\frac{7}{2})}^{x} = 2$ $(7/2)^x=2$
${3.5}^{x} = 2$ $3.5^x=2$

Using logarithms to solve, we arrive at

$x = {log}_{3.5} (2)$ $x=log_3.5(2)$

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How to find the x in this exponential function? Thank you!

$2 (7^{x} + 2^{x - 1}) = 7 (2^{x} - 7^{x - 1})$ $2(7^x+2^(x-1))=7(2^x-7^(x-1))$

1 Answer

Explanation:

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How to find the x in this exponential function? Thank you!

2(7x+2x−1)=7(2x−7x−1)2(7^x+2^(x-1))=7(2^x-7^(x-1))

1 Answer

Explanation:

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$2 (7^{x} + 2^{x - 1}) = 7 (2^{x} - 7^{x - 1})$ $2(7^x+2^(x-1))=7(2^x-7^(x-1))$