What is the answer to 2^{x+1}=5^{2x-1} ?

The answer to 2^{x+1}=5^{2x-1} is x=(ln(10))/(2ln(5)-ln(2))

Solve for x: 2^{x+1}=5^{2x-1}

The solution is x=(ln(10))/(2ln(5)-ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2^{x + 1} = 5^{2 x - 1}

x = \frac{ln ( 10 )}{2 ln ( 5 ) - ln ( 2 )}

Solution steps

2^{x + 1} = 5^{2 x - 1}

Apply exponent rules

(x + 1) ln (2) = (2 x - 1) ln (5)

Solve

(x + 1) ln (2) = (2 x - 1) ln (5) : x = \frac{ln ( 10 )}{2 ln ( 5 ) - ln ( 2 )}

x = \frac{ln ( 10 )}{2 ln ( 5 ) - ln ( 2 )}

3 2^{x} = 4

2^{x} = 2^{3}

\frac{2 ^{x}}{1 - 2 ^{x}} = 3

4 (\frac{2}{3})^{x^{2} - 5 x + 4} = 9

e^{k x} = 0

What is the answer to 2^{x+1}=5^{2x-1} ?
The answer to 2^{x+1}=5^{2x-1} is x=(ln(10))/(2ln(5)-ln(2))
Solve for x: 2^{x+1}=5^{2x-1}
The solution is x=(ln(10))/(2ln(5)-ln(2))