What is the answer to log_{10}(x^2)+7x=1 ?

The answer to log_{10}(x^2)+7x=1 is x=(2\W_{0}((7ln(10)e^{\frac{ln(10))/2})/(2))}{7ln(10)}

Solve for x: log_{10}(x^2)+7x=1

The solution is x=(2\W_{0}((7ln(10)e^{\frac{ln(10))/2})/(2))}{7ln(10)}

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lo g_{10} (x^{2}) + 7 x = 1

x = \frac{2 W _{0} ( \frac{7 l n ( 10 ) e ^{\frac{l n ( 10 )}{2}}}{2} )}{7 ln ( 10 )}

Decimal Notation

x = 0.29453 \dots

Solution steps

lo g_{10} (x^{2}) + 7 x = 1

Move

7 x

to the right side

lo g_{10} (x^{2}) = 1 - 7 x

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

1 0^{l o g_{10} (x^{2})} = 1 0^{1 - 7 x}

Simplify

1 0^{l o g_{10} (x^{2})} : x^{2}

x^{2} = 1 0^{1 - 7 x}

Solve

x^{2} = 1 0^{1 - 7 x} : x = \frac{2 W _{0} ( \frac{7 l n ( 10 ) e ^{\frac{l n ( 10 )}{2}}}{2} )}{7 ln ( 10 )}

x = \frac{2 W _{0} ( \frac{7 l n ( 10 ) e ^{\frac{l n ( 10 )}{2}}}{2} )}{7 ln ( 10 )}

Verify Solutions:

x = \frac{2 W _{0} ( \frac{7 l n ( 10 ) e ^{\frac{l n ( 10 )}{2}}}{2} )}{7 ln ( 10 )}

True

The solution is

x = \frac{2 W _{0} ( \frac{7 l n ( 10 ) e ^{\frac{l n ( 10 )}{2}}}{2} )}{7 ln ( 10 )}

3 lo g_{2} (x) - 1 = lo g_{2} (27)

lo g_{16} (x) = lo g_{4} (x^{2})

ln (x (x - 1)) = ln (x) + ln (x - 1)

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

lo g_{9} (4 x + 4) = 2

What is the answer to log_{10}(x^2)+7x=1 ?
The answer to log_{10}(x^2)+7x=1 is x=(2\W_{0}((7ln(10)e^{\frac{ln(10))/2})/(2))}{7ln(10)}
Solve for x: log_{10}(x^2)+7x=1
The solution is x=(2\W_{0}((7ln(10)e^{\frac{ln(10))/2})/(2))}{7ln(10)}