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

1 Answer
Sep 4, 2017

See explanation.

Explanation:

The equation is:

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

This can be transformed to:

6^(2x)+6*6^x-16=0

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

t^2+6t-16=0

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:

6^{x_1}=-8

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

6^{x_2}=2

x_2=log_6 2

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