What is the answer to log_{4}(3x+4)+log_{4}(16)=1 ?

The answer to log_{4}(3x+4)+log_{4}(16)=1 is x=-5/4

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log_{4}(3x+4)+log_{4}(16)=1

lo g_{4} (3 x + 4) + lo g_{4} (16) = 1

x = - \frac{5}{4}

Decimal Notation

x = - 1.25

Solution steps

lo g_{4} (3 x + 4) + lo g_{4} (16) = 1

Move

lo g_{4} (16)

to the right side

lo g_{4} (3 x + 4) = 1 - lo g_{4} (16)

Apply log rules

3 x + 4 = \frac{1}{4}

Solve

3 x + 4 = \frac{1}{4} : x = - \frac{5}{4}

x = - \frac{5}{4}

Verify Solutions:

x = - \frac{5}{4}

True

The solution is

x = - \frac{5}{4}

y, ln (y) = 2 t

ln (lo g_{10} (x^{2} + 1)) = 0

0 = 2 ln (x + 1)

x, ln (6 - x) = 0.6

ln (x) = \frac{ln ( x ) + 1}{2}

What is the answer to log_{4}(3x+4)+log_{4}(16)=1 ?
The answer to log_{4}(3x+4)+log_{4}(16)=1 is x=-5/4
Solve for x: log_{4}(3x+4)+log_{4}(16)=1
The solution is x=-5/4