What is the answer to log_{2}(x-1)+log_{2}(x+3)=1 ?

The answer to log_{2}(x-1)+log_{2}(x+3)=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_{2}(x-1)+log_{2}(x+3)=1

lo g_{2} (x - 1) + lo g_{2} (x + 3) = 1

x = - 1 + 6

Decimal Notation

x = 1.44948 \dots

Solution steps

lo g_{2} (x - 1) + lo g_{2} (x + 3) = 1

Apply log rules

(x - 1) (x + 3) = 2

Solve

(x - 1) (x + 3) = 2 : 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

lo g_{4} (2 - x) = lo g_{16} (13 - 4 x)

ln (x) = 10

lo g_{4} (x) = lo g_{4} (8 - x)

ln (3 x) + ln (x + 2) = 0

ln (x) + ln (x - 2) = ln (3)

What is the answer to log_{2}(x-1)+log_{2}(x+3)=1 ?
The answer to log_{2}(x-1)+log_{2}(x+3)=1 is x=-1+sqrt(6)
Solve for x: log_{2}(x-1)+log_{2}(x+3)=1
The solution is x=-1+sqrt(6)