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

The answer to e(x)=-log_{2}(10x) is x=(\W_{0}(\frac{ln(2)e)/(10))}{eln(2)}

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e(x)=-log_{2}(10x)

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

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

Decimal Notation

x = 0.08517 \dots

Solution steps

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

Move

lo g_{2} (10 x)

to the left side

e x + lo g_{2} (10 x) = 0

Move

e x

to the right side

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

f (x) = g (x),

then

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

2^{l o g_{2} (10 x)} = 2^{- e x}

Simplify

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

10 x = 2^{- e x}

Solve

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

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

Verify Solutions:

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

True

The solution is

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

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

lo g_{10} (x) = 15

lo g_{10} (x) = - 9

lo g_{x} (50) = 4

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

What is the answer to e(x)=-log_{2}(10x) ?
The answer to e(x)=-log_{2}(10x) is x=(\W_{0}(\frac{ln(2)e)/(10))}{eln(2)}