What is the answer to 8^{2x}+7(8^x)-18=0 ?

The answer to 8^{2x}+7(8^x)-18=0 is x= 1/3

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8^{2x}+7(8^x)-18=0

8^{2 x} + 7 (8^{x}) - 18 = 0

x = \frac{1}{3}

Solution steps

8^{2 x} + 7 (8^{x}) - 18 = 0

Apply exponent rules

(8^{x})^{2} + 7 \cdot 8^{x} - 18 = 0

Rewrite the equation with

8^{x} = u

(u)^{2} + 7 u - 18 = 0

Solve

u^{2} + 7 u - 18 = 0 : u = 2, u = - 9

u = 2, u = - 9

Substitute back

u = 8^{x},

solve for

x

Solve

8^{x} = 2 : x = \frac{1}{3}

Solve

8^{x} = - 9 :

No Solution for

x \in R

x = \frac{1}{3}

x, 2^{x} + 2^{y} + 2^{z} = e

e^{(7 - 4 x)} = 6

1 = (1 + 4.4 \cdot e^{- 0.737 B})

2^{3 x} - 1 = 0

25 (5^{2 x}) - 126 (5^{x}) + 5 = 0

What is the answer to 8^{2x}+7(8^x)-18=0 ?
The answer to 8^{2x}+7(8^x)-18=0 is x= 1/3
Solve for x: 8^{2x}+7(8^x)-18=0
The solution is x= 1/3