What is the answer to (x+1)=log_{10}(x-1)+3 ?

The answer to (x+1)=log_{10}(x-1)+3 is x=2,x=-(\W_{0}(-\frac{ln(10))/(10))-ln(10)}{ln(10)}

Solve for x: (x+1)=log_{10}(x-1)+3

The solution is x=2,x=-(\W_{0}(-\frac{ln(10))/(10))-ln(10)}{ln(10)}

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(x + 1) = lo g_{10} (x - 1) + 3

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

Solution steps

(x + 1) = lo g_{10} (x - 1) + 3

Switch sides

lo g_{10} (x - 1) + 3 = x + 1

Move

3

to the right side

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

f (x) = g (x),

then

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

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

Simplify

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

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

Solve

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

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

Verify Solutions:

x = 2

True

, x = - \frac{W _{0} ( - \frac{l n ( 10 )}{10} ) - ln ( 10 )}{ln ( 10 )}

True

The solutions are

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

lo g_{4} (x + 20) = 3

ln (x) - ln (x + 1) = 5

5 = lo g_{3} (x^{2} + 18)

ln (x - 1) - 1 = 0

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

What is the answer to (x+1)=log_{10}(x-1)+3 ?
The answer to (x+1)=log_{10}(x-1)+3 is x=2,x=-(\W_{0}(-\frac{ln(10))/(10))-ln(10)}{ln(10)}
Solve for x: (x+1)=log_{10}(x-1)+3
The solution is x=2,x=-(\W_{0}(-\frac{ln(10))/(10))-ln(10)}{ln(10)}