What is the answer to 64^x-8=7(4^x-2) ?

The answer to 64^x-8=7(4^x-2) is x=0,x= 1/2

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64^x-8=7(4^x-2)

6 4^{x} - 8 = 7 (4^{x} - 2)

x = 0, x = \frac{1}{2}

Decimal Notation

x = 0, x = 0.5

Solution steps

6 4^{x} - 8 = 7 (4^{x} - 2)

Apply exponent rules

(4^{x})^{3} - 8 = 7 (4^{x} - 2)

Rewrite the equation with

4^{x} = u

(u)^{3} - 8 = 7 (u - 2)

Solve

u^{3} - 8 = 7 (u - 2) : u = 1, u = 2, u = - 3

u = 1, u = 2, u = - 3

Substitute back

u = 4^{x},

solve for

x

Solve

4^{x} = 1 : x = 0

Solve

4^{x} = 2 : x = \frac{1}{2}

Solve

4^{x} = - 3 :

No Solution for

x \in R

x = 0, x = \frac{1}{2}

3^{x} = x^{2}

5 \cdot 1 0^{\frac{z}{4}} = 32

3^{2 x + 1} + 3 = 10 \cdot 3^{x}

e^{3 x + 1} = 100

5 = 6^{x}

What is the answer to 64^x-8=7(4^x-2) ?
The answer to 64^x-8=7(4^x-2) is x=0,x= 1/2
Solve for x: 64^x-8=7(4^x-2)
The solution is x=0,x= 1/2