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

The answer to l(x)=2-log_{10}(x-1) is x=(\W(\frac{ln(10)l)/(10^{-2+l)})+ln(10)l}{ln(10)l}

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l(x)=2-log_{10}(x-1)

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

x = \frac{W ( \frac{l n ( 10 ) l}{1 0 ^{- 2 + l}} ) + ln ( 10 ) l}{ln ( 10 ) l}

Solution steps

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

Move

lo g_{10} (x - 1)

to the left side

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

Move

l x

to the right side

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

f (x) = g (x),

then

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

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

Simplify

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

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

Solve

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

x = \frac{W ( \frac{l n ( 10 ) l}{1 0 ^{- 2 + l}} ) + ln ( 10 ) l}{ln ( 10 ) l}

Verify Solutions:

x = \frac{W ( \frac{l n ( 10 ) l}{1 0 ^{- 2 + l}} ) + ln ( 10 ) l}{ln ( 10 ) l}

True

The solution is

x = \frac{W ( \frac{l n ( 10 ) l}{1 0 ^{- 2 + l}} ) + ln ( 10 ) l}{ln ( 10 ) l}

0 = ln (0 + c)

8 lo g_{10} (x) = x

lo g_{q} (\frac{27}{64}) = - 3

3.4 = lo g_{10} (x)

- lo g_{10} (x) = 6.8

What is the answer to l(x)=2-log_{10}(x-1) ?
The answer to l(x)=2-log_{10}(x-1) is x=(\W(\frac{ln(10)l)/(10^{-2+l)})+ln(10)l}{ln(10)l}