How do you solve $3^(x+4)=7^(2x-1)$ ?

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Answer 1 · 2018-06-28T14:38:44

Take logarithms

$3^(x+4)=7^(2x-1)$

Take logs:

$(x+4)log3=(2x-1)log7$
$x(log3-2log7)=-log7-4log3$

$x=(-log7-4log3)/(log3-2log7)=(log7+4log3)/(2log7-log3)$
$=(log7+log81)/(log49-log3)=log567/log(49/3)=log_(49/3)567$

To three significant figures, $x=2.27$ .

How do you solve 3^(x+4)=7^(2x-1)3^(x+4)=7^(2x-1)?