How do you solve $6^{x} + 4^{x} = 9^{x}$ $6^x + 4^x = 9^x$ ?

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How do you solve $6^{x} + 4^{x} = 9^{x}$ $6^x + 4^x = 9^x$ ?

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Answer 1 · 2016-04-17T12:07:22

$x = \frac{ln (\frac{1 + \sqrt{5}}{2})}{ln (\frac{3}{2})}$ $x=(ln((1+sqrt(5))/2))/(ln (3/2))$

Explanation:

Divide by $4^{x}$ $4^x$ to form a quadratic in ${(\frac{3}{2})}^{x}$ $(3/2)^x$ .
Use $\frac{6^{x}}{4^{x}} = {(\frac{6}{4})}^{x} = {(\frac{3}{2})}^{x} and {(\frac{9}{4})}^{x} = {({(\frac{3}{2})}^{2})}^{x} = {({(\frac{3}{2})}^{x})}^{2}$ $6^x/4^x=(6/4)^x=(3/2)^x and (9/4)^x=((3/2)^2)^x=((3/2)^x)^2$ .
${({(\frac{3}{2})}^{x})}^{2} - {(\frac{3}{2})}^{x} - 1 = 0$ $((3/2)^x)^2-(3/2)^x-1=0$

So, ${(\frac{3}{2})}^{x} = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot (- 1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$ $(3/2)^x=(1+-sqrt(1-4*1*(-1)))/2=(1+-sqrt(5))/2$

For the positive solution:

${(\frac{3}{2})}^{x} = \frac{1 + \sqrt{5}}{2}$ $(3/2)^x=(1+sqrt(5))/2$

Applying logarythms:

$x ln (\frac{3}{2}) = ln (\frac{1 + \sqrt{5}}{2})$ $xln (3/2)=ln((1+sqrt(5))/2)$

$x=(ln((1+sqrt(5))/2))/(ln (3/2))=1.18681439....$

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How do you solve $6^{x} + 4^{x} = 9^{x}$ $6^x + 4^x = 9^x$ ?

1 Answer

Explanation:

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