What is the answer to log_{3}(x+4)+log_{3}(4x-7)=3 ?

The answer to log_{3}(x+4)+log_{3}(4x-7)=3 is x= 11/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

log_{3}(x+4)+log_{3}(4x-7)=3

lo g_{3} (x + 4) + lo g_{3} (4 x - 7) = 3

x = \frac{11}{4}

Decimal Notation

x = 2.75

Solution steps

lo g_{3} (x + 4) + lo g_{3} (4 x - 7) = 3

Apply log rules

(x + 4) (4 x - 7) = 27

Solve

(x + 4) (4 x - 7) = 27 : x = \frac{11}{4}, x = - 5

x = \frac{11}{4}, x = - 5

Verify Solutions:

x = \frac{11}{4}

True

, x = - 5

False

The solution is

x = \frac{11}{4}

0.81 = lo g_{10} (x)

lo g_{2} (3 \cdot 2^{x} - 4) = 2 x - 1

ln (4 x^{2}) + ln (x) = ln (32)

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

lo g_{3} (x) = lo g_{9} (x + 12)

What is the answer to log_{3}(x+4)+log_{3}(4x-7)=3 ?
The answer to log_{3}(x+4)+log_{3}(4x-7)=3 is x= 11/4
Solve for x: log_{3}(x+4)+log_{3}(4x-7)=3
The solution is x= 11/4