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

The answer to log_{4}(x+1)=2+log_{4}(3x-2) is x= 33/47

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

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

x = \frac{33}{47}

Decimal Notation

x = 0.70212 \dots

Solution steps

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

Apply log rules

x + 1 = 16 (3 x - 2)

Solve

x + 1 = 16 (3 x - 2) : x = \frac{33}{47}

x = \frac{33}{47}

Verify Solutions:

x = \frac{33}{47}

True

The solution is

x = \frac{33}{47}

lo g_{10} (x + 2) + lo g_{10} (4) = 1

7 - ln (3 - x) = 0

ln (x) + x = - 1

x, ln (x^{2}) = 14

lo g_{10} (∣ x ∣) = - 4

What is the answer to log_{4}(x+1)=2+log_{4}(3x-2) ?
The answer to log_{4}(x+1)=2+log_{4}(3x-2) is x= 33/47
Solve for x: log_{4}(x+1)=2+log_{4}(3x-2)
The solution is x= 33/47