Simple logs question

  • Thread starter Iclaudius
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  • #1
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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

  • #2
863
4
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.

[tex]\log_a (x+3) + \log_a (x-4) = \log_a (8)[/tex]

[tex]\log_a ((x+3)(x-4)) = \log_a (8)[/tex]

[tex]\log_a (x^2 - x -12) = \log_a (8)[/tex]

[tex]x^2 - x - 12 = 8[/tex]

So, what are the solutions for x? Now, plug them into the original equation and see which are valid.
 
Last edited:
  • #3
Zryn
Gold Member
310
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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.
 
  • #4
Mute
Homework Helper
1,388
10
There is one solution. You made a mistake when simplifying the logs. I'm going to guess you said something like

[tex]\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.[/tex]

Is that what you did? If so, that last implication is incorrect. Do you see why?
 
  • #5
36
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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
 
  • #6
36
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hmmm no...
 
  • #7
863
4
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.
 
  • #8
36
0
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
 
  • #9
863
4
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.

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

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

[tex]((x+3)(x-4)/8)=1[/tex]
 
Last edited:
  • #10
36
0
Ahhh - off course! Thank you so much for clearing that up!
 

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