logx+log(x+1)=log56. Solve the logarithmic equation.?

Recall that:

log(a) + log(b) = log(ab)

So we can combine these logs to get:

log[x(x + 1)] = log(56)

Since logs are single-valued for positive numbers, we can cancel them out to get:

x(x + 1) = 56

This is now a Quadratic Equation that we can easily solve.

x^2 + x = 56
==> x^2 + x - 56 = 0
==> (x + 8)(x - 7) = 0
==> x = -8 and x = 7

Checking shows that x = -8 produces a log of a negative number and that x = 7 is the only solution.

I hope this helps!

EDIT:

Actually, Morningfox, if you have a log of a negative number, it will have an attached πi at the end of the answer. Since you are ADDING two of them, the imaginary parts will NOT cancel out.

logx+log(x+1)=log56
logx(x+1) = log56
x(x+1) = 56
x^2+x = 56
x^2+x-56 =0
(x+8)(x-7) = 0
x=-8 or x=7

Hence x=7 is the only solution and ignore the solution x=-8 is because the logarithm of a negative number is not possible.

log x + log ( x+1 ) = log 56

log [ x ( x+1 ) ] = log 56

x ( x + 1 ) = 56

x ( x + 1 ) = 7 ( 7 + 1 ) so that x = 7. ........Ans.

logx(x+1) = log56 so x^2 + x = 56 then x^2 + x -56=0

factors (x + 8)(x -7)=0 then x + 8 = 0, x = -8


x - 7 = 0, x = 7

log(x) + log(x + 1) = log(56)
log[x(x + 1)] = log(56)
x(x + 1) = 56
x^2 + x - 56 = 0
x^2 + 8x - 7x - 56 = 0
(x^2 + 8x) - (7x + 56) = 0
x(x + 8) - 7(x + 8) = 0
(x + 8)(x - 7) = 0

x + 8 = 0
x = -8

x - 7 = 0
x = 7

∴ x = -8, 7

Actually, x = 7 or -8. The x = 7 part is easy to check. The x = -8 part needs complex logarithms, but it is also correct.


log(-8) = 2.079441542 + 3.141592654i
log(-7) = 1.945910149 - 3.141592654i
sum = 4.025351691
log(56) = 4.025351691