Question 313118
{{{3^(2x-1) = 10*(3^x)-3}}}
{{{3^(2x)/3=10*3^(x)-3}}}
Substitute {{{u=3^x}}}, then {{{u^2=3^(2x)}}}
{{{u^2/3=10*u-3}}}
{{{u^2=30u-9}}}
{{{u^2-30u+9=0}}}
Use the quadratic formula,
{{{u = (-(-30) +- sqrt( (-30)^2-4*1*9 ))/(2*1) }}}
{{{u = (30 +- sqrt( 900-36 ))/(2) }}}
{{{u = (30 +- sqrt( 864))/(2) }}}
{{{u = (30 +- 12sqrt( 6))/(2) }}}
{{{u = 15 +- 6sqrt( 6) }}}
Two solutions:
{{{u1=15+6sqrt(6)}}}
{{{3^(x)=15+6sqrt(6)}}}
{{{x=log(3,(15+6sqrt(6)))}}}
{{{x=log(10,(15+6sqrt(6)))/log(10,(3)) }}}
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{{{u2=15-6sqrt(6)}}}
{{{3^(x)=15-6sqrt(6)}}}
{{{x=log(3,(15-6sqrt(6)))}}}
{{{x=log(10,(15-6sqrt(6)))/log(10,(3)) }}}