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

The answer to log_{4}(3+x)=2+log_{4}(x+1) is x=-13/15

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

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

x = - \frac{13}{15}

Decimal Notation

x = - 0.86666 \dots

Solution steps

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

Apply log rules

3 + x = 16 (x + 1)

Solve

3 + x = 16 (x + 1) : x = - \frac{13}{15}

x = - \frac{13}{15}

Verify Solutions:

x = - \frac{13}{15}

True

The solution is

x = - \frac{13}{15}

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

lo g_{10} (\frac{x}{10}) = 2

5 ln (5 x + 9) - 4 = 11

0.44 = ln (x)

lo g_{10} (x) = \frac{4}{3}

What is the answer to log_{4}(3+x)=2+log_{4}(x+1) ?
The answer to log_{4}(3+x)=2+log_{4}(x+1) is x=-13/15
Solve for x: log_{4}(3+x)=2+log_{4}(x+1)
The solution is x=-13/15