Solve logaritmic equations

galwaystudent

I have an equation to solve for x: [latex]ln(x) + ln(x-1) = 1[/latex]

Is this correct -
[latex]ln(x)(x-1)=1[/latex]
[latex]e = (x)(x-1)[/latex]

Hence [latex]x = e, x = e+1[/latex]

MathsManiac

You can verify solution(s) to an equation by substituting the supposed solution(s) back into the original equation. In this case, that tells you that you're not correct.

The first two lines of your solution are correct, but the last line isn't. You can only do that last step when you have a product of factors equal to zero. The principle involved is that the only way for two things to multiply to give 0 is if one or other of them is 0. This is not true for other numbers.

What you have is a quadratic equation. You should get everything to one side, simplify, and then solve by factorising, by completing the square, or by using the -b formula.

gerry87

galwaystudent wrote: »

I have an equation to solve for x: [latex]ln(x) + ln(x-1) = 1[/latex]

Is this correct -
[latex]ln(x)(x-1)=1[/latex]
[latex]e = (x)(x-1)[/latex]

Hence [latex]x = e, x = e+1[/latex]

The thing you used in your last step only works when the equation is equal to 0.

for example you could have 3*4 = 12, 2*6 = 12 etc, where both multiplied numbers can be a few different things, whereas with x*y = 0, then either x or y has to be zero.

You can solve it with a quadratic equation so
[latex](x)(x-1) = e[/latex]
[latex]x^2 - x = e[/latex]
[latex]x^2 - x - e = 0[/latex]

now you use the formula
[latex]\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/latex]