What is the answer to log_{3}(5)+log_{3}(x-4)=2 ?

The answer to log_{3}(5)+log_{3}(x-4)=2 is x= 29/5

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log_{3}(5)+log_{3}(x-4)=2

lo g_{3} (5) + lo g_{3} (x - 4) = 2

x = \frac{29}{5}

Decimal

x = 5.8

Solution steps

lo g_{3} (5) + lo g_{3} (x - 4) = 2

Move

lo g_{3} (5)

to the right side

lo g_{3} (x - 4) = 2 - lo g_{3} (5)

Apply log rules

x - 4 = \frac{9}{5}

Solve

x - 4 = \frac{9}{5} : x = \frac{29}{5}

x = \frac{29}{5}

Verify Solutions:

x = \frac{29}{5}

True

The solution is

x = \frac{29}{5}

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

10 lo g_{10} (x) = 3

lo g_{8} (3 x - 2) = 2

2 \cdot lo g_{10} (3 - x) = - 1

lo g_{2} (8 + x) = 3

What is the answer to log_{3}(5)+log_{3}(x-4)=2 ?
The answer to log_{3}(5)+log_{3}(x-4)=2 is x= 29/5
Solve for x: log_{3}(5)+log_{3}(x-4)=2
The solution is x= 29/5