Solving Log Equation: Log_a(x+3) + Log_a(x-4) = Log_a(8)

[Iclaudius]

Hello friends,

so i have this problem it reads:

how many solutions does the following equation have

Log_a(x+3) + Log_a(x-4) = Log_a(8)

ok so first i put all the logs on one side of the equation and solve for x after using log laws,

and i get x = 4 and x = -3, however plugging these into the equation will result in Log(o) so i answer the question with the equation has 0 solutions but the book tells me it has 1 solution.

can someone please explain this to me - thank you very much for your time i do appreciate it,

Claudius

 

[KingNothing]

I think you just did some of your math wrong. You should not be getting 4 and -3 as solutions. I will update with some work in a moment.

\log_a (x+3) + \log_a (x-4) = \log_a (8)

\log_a ((x+3)(x-4)) = \log_a (8)

\log_a (x^2 - x -12) = \log_a (8)

x^2 - x - 12 = 8

So, what are the solutions for x? Now, plug them into the original equation and see which are valid.

 

[Zryn]

Using x = -3 you have log_a(0) + log_a(-7) = log_a(8), and using x = 4 you have log_a(7) + log_a(0) = log_a(8). Both these equations should tingle your funny bone!

There are 3 basic log laws governing addition, subtraction and powers, are you familiar with them?

Can you show your working so we can point out where you went wrong.

 

[Mute]

There is one solution. You made a mistake when simplifying the logs. I'm going to guess you said something like

\log_a(x+3)+\log_a(x-4) - \log_a(8) = 0 \Rightarrow \log_a((x+3)(x-4)/8) = 0 \Rightarrow (x+3)(x-4)/8 = 0.

Is that what you did? If so, that last implication is incorrect. Do you see why?

 

[Iclaudius]

ok so my working is like this

Ln( ((x+3) (x-4)) / 8 ) = 0

simple quadractic - solve for x i get x=4, and x =-3

 

[Iclaudius]

hmmm no...

 

[KingNothing]

I updated my original reply with some latex that might help you out. Sorry, I didn't realize PF Math was so poppin' at this hour.

 

[Iclaudius]

Ok i see what you did there King - however why can I not take Ln(8) to the other side and combine using log laws? As Mute has pointed out

 

[KingNothing]

Ok i see what you did there King - however why can I not take Ln(8) to the other side and combine using log laws? As Mute has pointed out

You can.

ok so my working is like this

Ln( ((x+3) (x-4)) / 8 ) = 0

simple quadractic - solve for x i get x=4, and x =-3

Your equation is correct - but remember that when you "undo" a logarithm, you effectively make each side of the equation the exponent of the base. So when you "undo" the log, the right side will have a^0, which is 1. Anything to the power of zero is one. Solving that quadratic will give you correct terms.

\log_a ((x+3)(x-4)/8)=0

a^{log_a ((x+3)(x-4)/8)}=a^0

((x+3)(x-4)/8)=1

 

[Iclaudius]

Ahhh - off course! Thank you so much for clearing that up!