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How to solve the equation 3^2x+1 = 3^x + 24, do I need logarithms

  • Thread starter repugno
  • Start date
  • #1
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Greetings all, I have a slight problem solving this equation ..... 3^2x+1 = 3^x + 24

I know I can solve it with logs but I'm just not sure where to begin. I have tried taking logarithms both sides but it didn't work :confused:

Perhaps someone could give me a clue :biggrin:

thanks
 

Answers and Replies

  • #2
Can you show the work you've done so far?

cookiemonster
 
  • #3
78
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Sure...

3^2x+1 = 3^x + 24

(2x+1)lg3 = xlg3 + lg24

(2x+1)lg3/lg3 = xlg3/lg3 + lg24/lg3

2x+1 = x + 2.89

x = 1.89

This is obviously wrong.
 
  • #4
Be careful with log (3^x + 24).

In general, log(a + b) does not equal log(a) + log(b).

And you need to be careful with your notation. I first read the problem as 3^(2x) + 1 = 3^x + 24. Those darned parentheses!

cookiemonster
 
  • #5
977
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32x+1 = 3x + 24

Can you show that 32x+1 = 3*32x? Can you show that 3*(32x) = 3*(3x)2? Now what kind of equation do you get if you let t = 3x?
 
  • #6
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Thank you very much for your help.
 
  • #7
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These logarithms is not my good friend unfortunitaly, it is either that or I'm missing a fundamental point.

[itex] log_3x - 2log_x3 = 1 [/itex]

I don't understand how to solve this, I know the rules of logarithms but they don't seem to be much help here. Can anyone help me please.

Thanks
 
  • #8
You might find this useful

[tex]\log_b a = \frac{\ln a}{\ln b}[/tex]

cookiemonster
 
  • #9
977
1
More generally:

[tex]\log_b a = \frac{\log_c a}{\log_c b}[/tex]

Where c is whatever number you want (cookiemonster chose c = e, resulting in natural logs). For the problem above you might want to pick c = 3.
 
  • #10
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It worked !! Thanks for the help yet again. One question though, how did you come to the conclusion that I should take the base as 3?
 
  • #11
977
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You started with this:

[itex] log_3x - 2log_x3 = 1 [/itex]

Using ln you would have gotten:

[itex] \frac{\ln x}{\ln 3} - 2\frac{\ln 3}{\ln x} = 1 [/itex]

And just get stuck with ln(3). However, using 3 as the base you can get:

[itex] \frac{\log_3 x}{\log_3 3} - 2\frac{\log_3 3}{\log_3 x} = 1 [/itex]

And of course log3(3) = 1 and it's much easier to handle:

[itex]\log_3 x - \frac{2}{\log_3 x} = 1 [/itex]

Obviously you would still get the same answer with both methods, I just thought it would be slightly easier to solve when using base 3. :smile:
 

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