How do you solve #6^ { 2x } + 6^ { x + 1} - 16= 0#?

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Answer 1 · 2017-09-04T12:09:14

See explanation.

The equation is:

#6^(2x)+6^(x+1)-16=0#

This can be transformed to:

Now if we replace #6^x# with a new variable #t=6^x# we get a quadratic equation:

This equation can be solved using the quadratic formula:

#Delta=6^2-4*1*(-16)=36+64=100#

#sqrt(Delta)=10#

#t_1=(-6-10)/2=-8#, #t_2=(-6+10)/2=2#

Now we have to find the values of #x# corresponding with calculated values of #t#:

This equation has no solution. There is no real value of #x# for which #6^x# is negative

#x_2=log_6 2#

The only solutin of the exponential equation is #x_2=log_6 2#