What is the answer to log_{e}(x^2)+log_{7}(x)=2 ?

The answer to log_{e}(x^2)+log_{7}(x)=2 is x=7^{(2log_{7}(e))/(2+log_{7)(e)}}

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log_{e}(x^2)+log_{7}(x)=2

lo g_{e} (x^{2}) + lo g_{7} (x) = 2

x = 7^{\frac{2 l o g _{7} ( e )}{2 + l o g _{7} ( e )}}

Decimal Notation

x = 2.21571 \dots

Solution steps

lo g_{e} (x^{2}) + lo g_{7} (x) = 2

Apply log rules

x = 7^{\frac{2 l o g _{7} ( e )}{2 + l o g _{7} ( e )}}

Verify Solutions:

x = 7^{\frac{2 l o g _{7} ( e )}{2 + l o g _{7} ( e )}}

True

The solution is

x = 7^{\frac{2 l o g _{7} ( e )}{2 + l o g _{7} ( e )}}

0.48 = lo g_{10} (x)

ln (y^{2} - 1) - ln (y + 1) = 3

x + lo g_{10} (x) = 0

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

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

What is the answer to log_{e}(x^2)+log_{7}(x)=2 ?
The answer to log_{e}(x^2)+log_{7}(x)=2 is x=7^{(2log_{7}(e))/(2+log_{7)(e)}}