How do I solve $3^{2 x} - 5 (3^{x}) = - 6$ $3^(2x) - 5(3^x)=-6$ ?

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Answer 1 · 2017-10-16T09:07:03

$x = 1$ $x=1$

or

$:.x=(log2)/(log3)~~0.6309297536$

$3^(2x)-5(3^x)=-6$

let $u=3^x$

then we have

$u^2-5u=-6$

$=>u^2-5u+6=0$

a quadratic in $u$

$(u-3)(u-2)=0$

$:.u=2, " or "u=3$

$u=3 "gives "3^x=3=>x=1$

$u=2 " gives "3^x=2$

$3^x=2$

take logs

$log(3^x)=log2$

$xlog3=log2$

$:.x=(log2)/(log3)~~0.6309297536$

How do I solve 3^(2x) - 5(3^x)=-632x−5(3x)=−63^(2x) - 5(3^x)=-6?