What is the answer to log_{x+1}(2x^2-7x+19)=2 ?

The answer to log_{x+1}(2x^2-7x+19)=2 is x=6,x=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

log_{x+1}(2x^2-7x+19)=2

lo g_{x + 1} (2 x^{2} - 7 x + 19) = 2

x = 6, x = 3

Solution steps

lo g_{x + 1} (2 x^{2} - 7 x + 19) = 2

Apply log rules

\frac{ln ( 2 x ^{2} - 7 x + 19 )}{ln ( x + 1 )} = 2

Multiply both sides by

ln (x + 1)

\frac{ln ( 2 x ^{2} - 7 x + 19 )}{ln ( x + 1 )} ln (x + 1) = 2 ln (x + 1)

Simplify

ln (2 x^{2} - 7 x + 19) = 2 ln (x + 1)

Apply log rules

2 x^{2} - 7 x + 19 = (x + 1)^{2}

Solve

2 x^{2} - 7 x + 19 = (x + 1)^{2} : x = 6, x = 3

x = 6, x = 3

Verify Solutions:

x = 6

True

, x = 3

True

The solutions are

x = 6, x = 3

ln (x) + ln (x + 12) = 4

ln (x) + ln (x + 12) = 3

ln (1 - (1)^{2} + 2 c) = 0

ln (x + 1) = 10

5 ln (x) + 3 ln (\frac{1}{x}) = 10

What is the answer to log_{x+1}(2x^2-7x+19)=2 ?
The answer to log_{x+1}(2x^2-7x+19)=2 is x=6,x=3
Solve for x: log_{x+1}(2x^2-7x+19)=2
The solution is x=6,x=3