What is the answer to log_{5}(x)+log_{5}(4x-1)=1 ?

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

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log_{5}(x)+log_{5}(4x-1)=1

lo g_{5} (x) + lo g_{5} (4 x - 1) = 1

x = \frac{5}{4}

Decimal Notation

x = 1.25

Solution steps

lo g_{5} (x) + lo g_{5} (4 x - 1) = 1

Apply log rules

x (4 x - 1) = 5

Solve

x (4 x - 1) = 5 : x = \frac{5}{4}, x = - 1

x = \frac{5}{4}, x = - 1

Verify Solutions:

x = \frac{5}{4}

True

, x = - 1

False

The solution is

x = \frac{5}{4}

ln (x) - ln (3) = 4

lo g_{2} (x) - lo g_{8} (x) = 4

ln (x) = lo g_{10} (x)

lo g_{10} (a^{2} - 15 a) = 2

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

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