[MOVED] solve ln(x-2) + ln(x-3) = ln (2x+24) for x

[tsh44]

Hi, the problem says to solve for x

ln(x-2) + ln(x-3) = ln (2x+24)

ln (x-2)(x-3) = ln (2x+4)
ln x^2-5x+6-(2x+4)=0

Take e^ of both sides and i got:

x^2-7x-11=0

I used the quadratic formula and got (7+SQRT(93))/2

I plugged this back into the equation and got slightly different answers on both sides. did i d something wrong or is that close enough? Thanks

 

[stapel]

Note: From your work shown, I will guess that the argument on the right-hand side is meant to be "2x + 4", not "2x + 24".

ln(x-2) + ln(x-3) = ln (2x+24)

ln (x-2)(x-3) = ln (2x+4)

At this stage, since you have "natural-log of (one thing) equals natural-log of (something else)", you can equate the arguments (that is, set the (one thing) equal to (something else)) and solve:

. . . . .x² - 5x + 6 = 2x + 4

. . . . .x² - 7x + 2 = 0

This doesn't factor, so you will need to use the Quadratic Formula. But the above should lead to a slightly different answer from what you'd gotten.

With respect to checking, I am assuming that you found the decimal approximations to x, and plugged those in. Due to round-off error, you will only get "close", even when correct (in the "exact" form of the solution). To be fairly sure of your work, try to use at least four decimal places on your approximations. That, or store the "exact" value in your calculator's memory, and plug that into the original equation. :idea:

Hope that helps!

Eliz.