What is the answer to log_{10}(6-2x)=1+log_{10}(x) ?

The answer to log_{10}(6-2x)=1+log_{10}(x) is x= 1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

log_{10}(6-2x)=1+log_{10}(x)

lo g_{10} (6 - 2 x) = 1 + lo g_{10} (x)

x = \frac{1}{2}

Decimal Notation

x = 0.5

Solution steps

lo g_{10} (6 - 2 x) = 1 + lo g_{10} (x)

Apply log rules

6 - 2 x = 10 x

Solve

6 - 2 x = 10 x : x = \frac{1}{2}

x = \frac{1}{2}

Verify Solutions:

x = \frac{1}{2}

True

The solution is

x = \frac{1}{2}

y, x = 2 lo g_{10} (y)

lo g_{10} (p^{- 1}) = 1

lo g_{3} (x) + 3 = 2

7.5 = 0.65 lo g_{10} (0 E) + 1.45

lo g_{10} (X) = 3.4146

What is the answer to log_{10}(6-2x)=1+log_{10}(x) ?
The answer to log_{10}(6-2x)=1+log_{10}(x) is x= 1/2
Solve for x: log_{10}(6-2x)=1+log_{10}(x)
The solution is x= 1/2