What is the answer to log_{10}(5x+1)=2+log_{10}(2x-3) ?

The answer to log_{10}(5x+1)=2+log_{10}(2x-3) is x= 301/195

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log_{10}(5x+1)=2+log_{10}(2x-3)

lo g_{10} (5 x + 1) = 2 + lo g_{10} (2 x - 3)

x = \frac{301}{195}

Decimal Notation

x = 1.54358 \dots

Solution steps

lo g_{10} (5 x + 1) = 2 + lo g_{10} (2 x - 3)

Apply log rules

5 x + 1 = 100 (2 x - 3)

Solve

5 x + 1 = 100 (2 x - 3) : x = \frac{301}{195}

x = \frac{301}{195}

Verify Solutions:

x = \frac{301}{195}

True

The solution is

x = \frac{301}{195}

4 lo g_{2} (2 x) = 36

lo g_{10} (n) = - 0.35

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

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

lo g_{10} (10 x) - lo g_{10} (4 - x) = 2

What is the answer to log_{10}(5x+1)=2+log_{10}(2x-3) ?
The answer to log_{10}(5x+1)=2+log_{10}(2x-3) is x= 301/195
Solve for x: log_{10}(5x+1)=2+log_{10}(2x-3)
The solution is x= 301/195