What is the answer to solvefor x,2^x-2^y=128 ?

The answer to solvefor x,2^x-2^y=128 is x=(ln(128+2^y))/(ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor x,2^x-2^y=128

solve for

x, 2^{x} - 2^{y} = 128

x = \frac{ln ( 128 + 2 ^{y} )}{ln ( 2 )}

Solution steps

2^{x} - 2^{y} = 128

Move

2^{y}

to the right side

2^{x} = 128 + 2^{y}

Apply exponent rules

x ln (2) = ln (128 + 2^{y})

Solve

x ln (2) = ln (128 + 2^{y}) : x = \frac{ln ( 128 + 2 ^{y} )}{ln ( 2 )}

x = \frac{ln ( 128 + 2 ^{y} )}{ln ( 2 )}

6 = 12 \cdot 5^{- k^{20}}

7^{- x} \cdot 7^{- x} = 343

0 = 10 e^{0.002 x}

6^{4 x - 4} = 1

e^{x^{2}} = e^{x} (e^{x - 1})^{2}

What is the answer to solvefor x,2^x-2^y=128 ?
The answer to solvefor x,2^x-2^y=128 is x=(ln(128+2^y))/(ln(2))