What is the answer to 16^x-5(4^{x+1})+64=0 ?

The answer to 16^x-5(4^{x+1})+64=0 is x=2,x=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

16^x-5(4^{x+1})+64=0

1 6^{x} - 5 (4^{x + 1}) + 64 = 0

x = 2, x = 1

Solution steps

1 6^{x} - 5 (4^{x + 1}) + 64 = 0

Apply exponent rules

(4^{x})^{2} - 5 \cdot 4^{x} \cdot 4^{1} + 64 = 0

Rewrite the equation with

4^{x} = u

(u)^{2} - 5 u \cdot 4^{1} + 64 = 0

Solve

u^{2} - 5 u \cdot 4^{1} + 64 = 0 : u = 16, u = 4

u = 16, u = 4

Substitute back

u = 4^{x},

solve for

x

Solve

4^{x} = 16 : x = 2

Solve

4^{x} = 4 : x = 1

x = 2, x = 1

2^{3 x + 1} = 8

(8^{9})^{p} = 8^{18}

x, y e^{x} = 1

4 (2^{x}) = 3^{x} \cdot 5^{x}

\frac{e ^{x} + e ^{- x}}{2} = 1

What is the answer to 16^x-5(4^{x+1})+64=0 ?
The answer to 16^x-5(4^{x+1})+64=0 is x=2,x=1
Solve for x: 16^x-5(4^{x+1})+64=0
The solution is x=2,x=1