What is the answer to log_{10}(x)=-x+40 ?

The answer to log_{10}(x)=-x+40 is x=(\W_{0}(ln(10)*10^{40}))/(ln(10))

Solve for x: log_{10}(x)=-x+40

The solution is x=(\W_{0}(ln(10)*10^{40}))/(ln(10))

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lo g_{10} (x) = - x + 40

x = \frac{W _{0} ( ln ( 10 ) \cdot 1 0 ^{40} )}{ln ( 10 )}

Solution steps

lo g_{10} (x) = - x + 40

f (x) = g (x),

then

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

1 0^{l o g_{10} (x)} = 1 0^{- x + 40}

Simplify

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

x = 1 0^{- x + 40}

Solve

x = 1 0^{- x + 40} : x = \frac{W _{0} ( ln ( 10 ) \cdot 1 0 ^{40} )}{ln ( 10 )}

x = \frac{W _{0} ( ln ( 10 ) \cdot 1 0 ^{40} )}{ln ( 10 )}

Verify Solutions:

x = \frac{W _{0} ( ln ( 10 ) \cdot 1 0 ^{40} )}{ln ( 10 )}

True

The solution is

x = \frac{W _{0} ( ln ( 10 ) \cdot 1 0 ^{40} )}{ln ( 10 )}

ln (1 - 2 x) = ln (2 - x)

lo g_{3} (5 x - 7) = 0

0 = 5 (ln (4 x) + 1)

ln (x) = ln (x + 6) - ln (x - 4)

x, y x = - lo g_{10} (x)

What is the answer to log_{10}(x)=-x+40 ?
The answer to log_{10}(x)=-x+40 is x=(\W_{0}(ln(10)*10^{40}))/(ln(10))
Solve for x: log_{10}(x)=-x+40
The solution is x=(\W_{0}(ln(10)*10^{40}))/(ln(10))