What is the answer to log_{10}(2x)+log_{10}(x)=log_{10}(5)+1 ?

The answer to log_{10}(2x)+log_{10}(x)=log_{10}(5)+1 is x=10^{(log_{10}(5)+1-log_{10}(2))/2}

Solve for x: log_{10}(2x)+log_{10}(x)=log_{10}(5)+1

The solution is x=10^{(log_{10}(5)+1-log_{10}(2))/2}

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lo g_{10} (2 x) + lo g_{10} (x) = lo g_{10} (5) + 1

x = 1 0^{\frac{l o g _{10} ( 5 ) + 1 - l o g _{10} ( 2 )}{2}}

Decimal

x = 5

Solution steps

lo g_{10} (2 x) + lo g_{10} (x) = lo g_{10} (5) + 1

Apply log rules

x = 1 0^{\frac{l o g _{10} ( 5 ) + 1 - l o g _{10} ( 2 )}{2}}

Verify Solutions:

x = 1 0^{\frac{l o g _{10} ( 5 ) + 1 - l o g _{10} ( 2 )}{2}}

True

The solution is

x = 1 0^{\frac{l o g _{10} ( 5 ) + 1 - l o g _{10} ( 2 )}{2}}

ln (x) + ln (x + 1) = 4

lo g_{\frac{1}{6}} (3 - 10 x) = - 2

lo g_{10} (3 x + 1) - lo g_{10} (2 x - 3) = 1 - lo g_{10} (5)

5 x^{4} ln (x) + x^{4} = 0

\frac{ln ( 6 b )}{6} = 0

What is the answer to log_{10}(2x)+log_{10}(x)=log_{10}(5)+1 ?
The answer to log_{10}(2x)+log_{10}(x)=log_{10}(5)+1 is x=10^{(log_{10}(5)+1-log_{10}(2))/2}
Solve for x: log_{10}(2x)+log_{10}(x)=log_{10}(5)+1
The solution is x=10^{(log_{10}(5)+1-log_{10}(2))/2}