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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = - 1 + 6

Decimal Notation

x = 1.44948 \dots

Solution steps

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

Apply log rules

(x + 2) \cdot 2 x = 10

Solve

(x + 2) \cdot 2 x = 10 : x = - 1 + 6, x = - 1 - 6

x = - 1 + 6, x = - 1 - 6

Verify Solutions:

x = - 1 + 6

True

, x = - 1 - 6

False

The solution is

x = - 1 + 6

L (v) = v ln (1 - v^{2})

lo g_{b} (81) = 2

lo g_{10} (x^{2} - 16 x + 1) = 0

2 ln (3 x^{2} + 1) = 13

ln (15 - 2 x) = 2 ln (x)

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