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

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Answer 1 · 2016-04-17T12:07:22

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

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

So, $(3/2)^x=(1+-sqrt(1-4*1*(-1)))/2=(1+-sqrt(5))/2$

For the positive solution:

$(3/2)^x=(1+sqrt(5))/2$

Applying logarythms:

$xln (3/2)=ln((1+sqrt(5))/2)$

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

How do you solve 6^x + 4^x = 9^x6^x + 4^x = 9^x?