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

The answer to log_{5}(x+1)=2+log_{5}(3x-2) is x= 51/74

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

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

x = \frac{51}{74}

Decimal Notation

x = 0.68918 \dots

Solution steps

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

Apply log rules

x + 1 = 25 (3 x - 2)

Solve

x + 1 = 25 (3 x - 2) : x = \frac{51}{74}

x = \frac{51}{74}

Verify Solutions:

x = \frac{51}{74}

True

The solution is

x = \frac{51}{74}

ln ∣ x ∣ = ln (x^{2})

lo g_{4} (v) = 4

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

ln (x + 1) = ln (x) ln (1)

lo g_{25} (x) = \frac{3}{2}

What is the answer to log_{5}(x+1)=2+log_{5}(3x-2) ?
The answer to log_{5}(x+1)=2+log_{5}(3x-2) is x= 51/74
Solve for x: log_{5}(x+1)=2+log_{5}(3x-2)
The solution is x= 51/74