What is the answer to solvefor x,antilog_{10}(25)=32^x ?

The answer to solvefor x,antilog_{10}(25)=32^x is x=(ln(2log_{10}(5)anit))/(5ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor x,antilog_{10}(25)=32^x

solve for

x, an t i lo g_{10} (25) = 3 2^{x}

x = \frac{ln ( 2 lo g _{10} ( 5 ) ani t )}{5 ln ( 2 )}

Solution steps

an t i lo g_{10} (25) = 3 2^{x}

Apply exponent rules

ln (an t i lo g_{10} (25)) = x ln (32)

Solve

ln (an t i lo g_{10} (25)) = x ln (32) : x = \frac{ln ( 2 lo g _{10} ( 5 ) ani t )}{5 ln ( 2 )}

x = \frac{ln ( 2 lo g _{10} ( 5 ) ani t )}{5 ln ( 2 )}

x, 5^{x + 1} = 10

4^{x - 8} = 8

4^{x - 8} = 6

2^{2 x + 1} - 2 x - 6 = 0

3^{x - 1} + 3^{x} + 3^{x + 1} = 39

What is the answer to solvefor x,antilog_{10}(25)=32^x ?
The answer to solvefor x,antilog_{10}(25)=32^x is x=(ln(2log_{10}(5)anit))/(5ln(2))