# Simple logs question

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

## Answers and Replies

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.

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Zryn
Gold Member
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
Homework Helper
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?

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

hmmm no...

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.

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

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.

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

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

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

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