What is the answer to log_{4}(10)+log_{4}(2x-1)=1 ?

The answer to log_{4}(10)+log_{4}(2x-1)=1 is x= 7/10

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log_{4}(10)+log_{4}(2x-1)=1

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

x = \frac{7}{10}

Decimal Notation

x = 0.7

Solution steps

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

Move

lo g_{4} (10)

to the right side

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

Apply log rules

2 x - 1 = \frac{2}{5}

Solve

2 x - 1 = \frac{2}{5} : x = \frac{7}{10}

x = \frac{7}{10}

Verify Solutions:

x = \frac{7}{10}

True

The solution is

x = \frac{7}{10}

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

lo g_{10} (\frac{800}{x}) = 0.3

lo g_{2} (n + 1) = 10

1 = ln (ln (x))

5 lo g_{10} (x) = 2 lo g_{10} (x)

What is the answer to log_{4}(10)+log_{4}(2x-1)=1 ?
The answer to log_{4}(10)+log_{4}(2x-1)=1 is x= 7/10
Solve for x: log_{4}(10)+log_{4}(2x-1)=1
The solution is x= 7/10