How do you solve $4 (2^{x}) = 5^{x}$ $4(2^x) = 5^x$ ?

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How do you solve $4 (2^{x}) = 5^{x}$ $4(2^x) = 5^x$ ?

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Answer 1 · 2015-11-18T01:46:36

I found: $x = 1.5129$ $x=1.5129$

Explanation:

Start by writing it as:
$2^{2} \cdot 2^{x} = 5^{x}$ $2^2*2^x=5^x$
$2^{x + 2} = 5^{x}$ $2^(x+2)=5^x$
take the $ln$ $ln$ of both sides:
$ln (2^{x + 2}) = {ln 5}^{x}$ $ln(2^(x+2))=ln5^x$
use the property of logs:
${log x}^{a} = a log x$ $logx^a=alogx$
to get:
$(x + 2) ln 2 = x ln 5$ $(x+2)ln2=xln5$
$x ln 2 + 2 ln 2 - x ln 5 = 0$ $xln2+2ln2-xln5=0$
isolate $x$ $x$ :
$x (ln 2 - ln 5) = - 2 ln 2$ $x(ln2-ln5)=-2ln2$
$x = - \frac{2 ln 2}{ln 2 - ln 5} = - \frac{1.3863}{- 0.9163} = 1.5129$ $x=-(2ln2)/(ln2-ln5)=-1.3863/-0.9163=1.5129$

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How do you solve $4 (2^{x}) = 5^{x}$ $4(2^x) = 5^x$ ?

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