What is the answer to log_{2}(x-1)=4-log_{2}(x+1) ?

The answer to log_{2}(x-1)=4-log_{2}(x+1) is x=sqrt(17)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

log_{2}(x-1)=4-log_{2}(x+1)

lo g_{2} (x - 1) = 4 - lo g_{2} (x + 1)

x = 17

Decimal Notation

x = 4.12310 \dots

Solution steps

lo g_{2} (x - 1) = 4 - lo g_{2} (x + 1)

Add

lo g_{2} (x + 1)

to both sides

lo g_{2} (x - 1) + lo g_{2} (x + 1) = 4 - lo g_{2} (x + 1) + lo g_{2} (x + 1)

Simplify

lo g_{2} (x - 1) + lo g_{2} (x + 1) = 4

Apply log rules

(x - 1) (x + 1) = 16

Solve

(x - 1) (x + 1) = 16 : x = 17, x = - 17

x = 17, x = - 17

Verify Solutions:

x = 17

True

, x = - 17

False

The solution is

x = 17

lo g_{6} (3 x^{2} - 3 x) = 2

lo g_{10} (x - 3) 9 = 2

lo g_{3} (6 x) + lo g_{3} (2) = 3

x, ln (y) = - 3 ln (x) + c

lo g_{3} (4 x + 10) = lo g_{3} (x + 1)

What is the answer to log_{2}(x-1)=4-log_{2}(x+1) ?
The answer to log_{2}(x-1)=4-log_{2}(x+1) is x=sqrt(17)
Solve for x: log_{2}(x-1)=4-log_{2}(x+1)
The solution is x=sqrt(17)