# Solve for x (4*16)^x+(4*2)^x=79.

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To solve for x:

(4*16)^x+(4*2)^x=79.

Solution:

Given equation is rewritten as: 64^x +8^x = 79.

Let 64^x = (8^x)^2 = t^2, where t = 8^x.

Therefore the equation becomes: t^2+t - 79 = 0

t^2 +t -79 = 0.

We use quadratic formula for solutions:

t1 = {-1+ sqrt(1+4*79)}/2 and t2 = {-1-sqqr(1+4*79)}/2

t1 = {-1+sqrt 317}/2 =

8^x = (-1+sqrt317)/2.

We take logarithms:

xlog8 = log(-1+log317)/2}

**Therefore x= log{(1+sqrt317)/2 }/log8.**** **

We'll calculate the multiplication inside the first pair of brackets.

4*16 = 64

We'll calculate the multiplication inside the second pair of brackets.

4*2 = 8

We notice that 64 = 8^2

We'll raise to x power both sides:

64^x = 8^2x

We'll re-write the equation:

8^2x + 8^x = 79

We'll substitute 8^x = t

t^2 + t - 79 = 0

We'll apply quadratic formula:

t1 = [-1+sqrt(1+316)]/2

t1 = (-1+17.8044)/2 approx.

t1 = 16.8044/2

t1 = 8.4022

8^x = t1

8^x = 8.4022

We'll take logarithms both sides:

ln 8^x = ln 8.4022

x = ln 8.4022/ln 8

t2 = (-1-17.8044)/2 approx.

t2 = -9.4022

8^x = -9.4022 undefined!

**The solution of the equation is x = ln 8.4022/ln 8 approx.**