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

The answer to log_{4}(2x+1)+log_{4}(x-2)=1 is x=(3+sqrt(57))/4

Solve for x: log_{4}(2x+1)+log_{4}(x-2)=1

The solution is x=(3+sqrt(57))/4

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

x = \frac{3 + 57}{4}

Decimal Notation

x = 2.63745 \dots

Solution steps

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

Apply log rules

(2 x + 1) (x - 2) = 4

Solve

(2 x + 1) (x - 2) = 4 : x = \frac{3 + 57}{4}, x = \frac{3 - 57}{4}

x = \frac{3 + 57}{4}, x = \frac{3 - 57}{4}

Verify Solutions:

x = \frac{3 + 57}{4}

True

, x = \frac{3 - 57}{4}

False

The solution is

x = \frac{3 + 57}{4}

0.69 = ln (x)

x + ln (x) = e + 1

5 ln (5 x - 4) - 15 = 15

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

lo g_{2} (x) \cdot ln (2) = 1

What is the answer to log_{4}(2x+1)+log_{4}(x-2)=1 ?
The answer to log_{4}(2x+1)+log_{4}(x-2)=1 is x=(3+sqrt(57))/4
Solve for x: log_{4}(2x+1)+log_{4}(x-2)=1
The solution is x=(3+sqrt(57))/4