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

[repugno]

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

Perhaps someone could give me a clue

thanks

 

[cookiemonster]

Can you show the work you've done so far?

cookiemonster

 

[repugno]

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.

 

[cookiemonster]

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

 

[Chen]

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

Can you show that 3^2x+1 = 3*3^2x? Can you show that 3*(3^2x) = 3*(3^x)²? Now what kind of equation do you get if you let t = 3^x?

 

[repugno]

Thank you very much for your help.

 

[repugno]

These logarithms is not my good friend unfortunitaly, it is either that or I'm missing a fundamental point.

log_3x - 2log_x3 = 1

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

 

[cookiemonster]

You might find this useful

\log_b a = \frac{\ln a}{\ln b}

cookiemonster

 

[Chen]

More generally:

\log_b a = \frac{\log_c a}{\log_c b}

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.

 

[repugno]

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?

 

[Chen]

You started with this:

log_3x - 2log_x3 = 1

Using ln you would have gotten:

\frac{\ln x}{\ln 3} - 2\frac{\ln 3}{\ln x} = 1

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

\frac{\log_3 x}{\log_3 3} - 2\frac{\log_3 3}{\log_3 x} = 1

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

\log_3 x - \frac{2}{\log_3 x} = 1

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.