help with some questions (solving logarithmic equations)

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
trdert
Posts: 6
Joined: Tue Aug 04, 2009 1:21 pm
Contact:

help with some questions (solving logarithmic equations)

Postby trdert » Tue Aug 04, 2009 1:29 pm

hi,

i'm trying to bone up on my pre-calc and calc for a placement exam in a couple weeks. the school gave me some guidance on review and one of the sites they reference is "calculus on the web" which is supported by temple univ. in the logarithmic equations section of their pre-calculus book i came across three problems i can't seem to get the correct solutions for:

#6) ln (9 - x2) = 2
#10) ln (-3x + 1) = ln 2x + 3
and #15) ln (x3 - 2) = ln ( 4x2 - 5x)

if anyone out there could lend a hand it would be most appreciated. thanks very much in advance.

tim

User avatar
stapel_eliz
Posts: 1738
Joined: Mon Dec 08, 2008 4:22 pm
Contact:

Postby stapel_eliz » Tue Aug 04, 2009 4:15 pm

trdert wrote:#6) ln (9 - x2) = 2

Use The Relationship to convert this log equation to the corresponding exponential form.

. . . . .

Then solve the resulting quadratic by taking square roots.

trdert wrote:#10) ln (-3x + 1) = ln 2x + 3

Subtract the ln(2x) over to the left-hand side, apply a log rule to combine the two logs into one, and then convert the resulting log equation to the corresponding exponential equation.

. . . . .

Then solve the resulting rational equation.

trdert wrote:#15) ln (x3 - 2) = ln ( 4x2 - 5x)

Since this is already in the form "log(something) equals log(something else)", just set the arguments equal. Then solve the resulting cubic equation.

If you get stuck, please reply showing how far you have gotten. Thank you! :D

trdert
Posts: 6
Joined: Tue Aug 04, 2009 1:21 pm
Contact:

Re: help with some questions (solving logarithmic equations)

Postby trdert » Tue Aug 04, 2009 7:17 pm

awesome, thanks. so i think i understand the first two but for the last one i'm still hung up.

put all the terms on one side and set equal to "0" but i'm getting hung up on the rational roots test. -2 is the constant and 1 is the leading coefficient. so that leaves a possible solution set of +/-2 and +/-1. i see why 2 is a solution and why -2 and -1 are not but why is 1 not a solution?

i.e., plugging it into the equation: (1)3 - 4(1)2+5(1)-2 = 0, right?

the website (calculus on the web at temple is indicating that 2 is a solution but 1 is not).

thanks so much for your help!

User avatar
stapel_eliz
Posts: 1738
Joined: Mon Dec 08, 2008 4:22 pm
Contact:

Postby stapel_eliz » Tue Aug 04, 2009 8:24 pm

Keep in mind how logs are defined. If x = 1, what value do you get inside the logs? In particular, what is the sign?

Is this allowed, for logs? :wink:

trdert
Posts: 6
Joined: Tue Aug 04, 2009 1:21 pm
Contact:

Re: help with some questions (solving logarithmic equations)

Postby trdert » Wed Aug 05, 2009 8:49 pm

ah, logs aren't defined for negatives...


Return to “Advanced Algebra ("pre-calculus")”