What is the answer to log_{3}(x-2)+log_{3}(4x-7)=1 ?

The answer to log_{3}(x-2)+log_{3}(4x-7)=1 is x= 11/4

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log_{3}(x-2)+log_{3}(4x-7)=1

lo g_{3} (x - 2) + lo g_{3} (4 x - 7) = 1

x = \frac{11}{4}

Decimal Notation

x = 2.75

Solution steps

lo g_{3} (x - 2) + lo g_{3} (4 x - 7) = 1

Apply log rules

(x - 2) (4 x - 7) = 3

Solve

(x - 2) (4 x - 7) = 3 : x = \frac{11}{4}, x = 1

x = \frac{11}{4}, x = 1

Verify Solutions:

x = \frac{11}{4}

True

, x = 1

False

The solution is

x = \frac{11}{4}

ln (x - 8) + ln (x + 1) = 0

10 \cdot lo g_{10} (4 x) = 25

x, 3 ln (3 x + 9) + 2 = 17

lo g_{10} (x) = 47712

lo g_{10} (\frac{24 - X}{X}) = 456

What is the answer to log_{3}(x-2)+log_{3}(4x-7)=1 ?
The answer to log_{3}(x-2)+log_{3}(4x-7)=1 is x= 11/4
Solve for x: log_{3}(x-2)+log_{3}(4x-7)=1
The solution is x= 11/4