en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5^2=25inlog_{10}(x)

Solution

5^{2} = 25 in lo g_{10} (x)

Solution

x = 1 0^{\frac{1}{in}}

Solution steps

5^{2} = 25 in lo g_{10} (x)

Switch sides

25 in lo g_{10} (x) = 5^{2}

5^{2} = 25

25 in lo g_{10} (x) = 25

Divide both sides by

25 in

lo g_{10} (x) = \frac{1}{in}

Apply log rules

x = 1 0^{\frac{1}{in}}

Popular Examples

log_{10}(z)+12=9

lo g_{10} (z) + 12 = 9

solvefor x,ln(y)=ln(x+1)solve for

x, ln (y) = ln (x + 1)

x=1+3.322(log_{28}(x))

x = 1 + 3.322 (lo g_{28} (x))

log_{10}(x)=1.97-1.32

lo g_{10} (x) = 1.97 - 1.32

solvefor x,ln(y)=ln(x+2)solve for

x, ln (y) = ln (x + 2)

Frequently Asked Questions (FAQ)

What is the answer to 5^2=25inlog_{10}(x) ?
The answer to 5^2=25inlog_{10}(x) is x=10^{1/(in)}