What is the answer to 4^{x+2}-5(4^x)=99 ?

The answer to 4^{x+2}-5(4^x)=99 is x=(ln(3))/(ln(2))

Solve for x: 4^{x+2}-5(4^x)=99

The solution is x=(ln(3))/(ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4^{x + 2} - 5 (4^{x}) = 99

x = \frac{ln ( 3 )}{ln ( 2 )}

Decimal Notation

x = 1.58496 \dots

Solution steps

4^{x + 2} - 5 (4^{x}) = 99

Factor

4^{x + 2} - 5 \cdot 4^{x} : 11 \cdot 4^{x}

11 \cdot 4^{x} = 99

Divide both sides by

11

4^{x} = 9

Apply exponent rules

x ln (4) = ln (9)

Solve

x ln (4) = ln (9) : x = \frac{ln ( 3 )}{ln ( 2 )}

x = \frac{ln ( 3 )}{ln ( 2 )}

2 5^{3 x} = 12 5^{4 x - 2}

x, 2^{3 - x} = 5^{x}

5^{x + 1} + 5 x = 6

4 (3^{x + 1}) = 24

4 (3^{x}) = 3 (6^{(3 x - 3)})

What is the answer to 4^{x+2}-5(4^x)=99 ?
The answer to 4^{x+2}-5(4^x)=99 is x=(ln(3))/(ln(2))
Solve for x: 4^{x+2}-5(4^x)=99
The solution is x=(ln(3))/(ln(2))