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

The answer to log_{3}(2x-3)+log_{3}(5)=4 is x= 48/5

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

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

x = \frac{48}{5}

Solution steps

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

Move

lo g_{3} (5)

to the right side

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

Apply log rules

2 x - 3 = \frac{81}{5}

Solve

2 x - 3 = \frac{81}{5} : x = \frac{48}{5}

x = \frac{48}{5}

Verify Solutions:

x = \frac{48}{5}

True

The solution is

x = \frac{48}{5}

lo g_{6} (x + 1) + lo g_{6} (2) = 2

2 + ln (x + 5) = 7

a (t) lo g_{10} (t) = a (t) y

2.38 = - lo g_{10} (x)

lo g_{10} (x) + lo g_{10} (x + 5) = 4

What is the answer to log_{3}(2x-3)+log_{3}(5)=4 ?
The answer to log_{3}(2x-3)+log_{3}(5)=4 is x= 48/5
Solve for x: log_{3}(2x-3)+log_{3}(5)=4
The solution is x= 48/5