How do you solve $4^(3-2x)=5^-x$ ?

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Answer 1 · 2017-07-26T19:52:24

$x=((3log4)/(2log4-log5))≈3.5755$

$4^(3-2x)=5^-x$

Take the common logarithm of both sides
$log4^(3-2x) = log5^-x$
Move the exponents in front of the logarithms
$(3-2x)*log4 = -xlog5$
Distribute
$3log4-2xlog4=-xlog5$
Rearrange the terms so that each "x" appears on the same side of the equation
$3log4=2xlog4-xlog5$
Factor out x from the right side of the equation
$3log4=x(2log4-log5)$
Isolate (solve for) the variable x by dividing both sides of the equation by $(2log4-log5)$ .
$(3log4)/(2log4-log5)=(xcancel((2log4-log5)))/(cancel(2log4-log5))$
(Optional) Plug into calculator to approximate a value for x
$x=(3log4)/(2log4-log5)≈3.5755$

How do you solve 4^(3-2x)=5^-x4^(3-2x)=5^-x?