Solving Logarithm Equations: Tips and Hints | TIA

[swears]

Hey guys, I got stuck doing these HW problems and I was hoping someone could give me some hints on how to solve them.

1.) 4 * 3^(X) = 7 * 5^(X)

2.) 2^(X) = e^(X+1)

Any help would be appreciated.
TIA

 

[Hootenanny]

Hi 'swears' and welcome to PF,

Could you please show what you have attempted thus far?

~H

 

[swears]

Ok, well I know I need to use the 'log' function since we are using exponents for the first problem, and the "ln" function for the second. I learned how to solve 1 variable exponents in class today, but since these have 2 I'm not sure how to approach them.

I could rewrite all the attempts I made on the problems, but none of them got me close to the right answer, and they are getting difficult to read after the numerous times I erased them.

 

[TD]

For the first one: try to get everything with an unknown exponent to the left side, and the rest to the right side. Can you 'group' the expression at the left side, so there's only one exponent? Use the properties of powers. Then take the logarithm of both sides.

 

[swears]

Ok, thanks TD. I just realized something, but I'm not sure if I did it right.

I ended up with 3^(X) - 5^(X) = 1.75

Am I allowed to subtract those to get -2^X = 1.75?

 

[TD]

Oh no, watch out! You have a product: they're all factors, not terms.

If you have a*x = c*y and you want x and y on the same side, you get: x/y = c/a.
Remember: you're allowed to multiply both sides with a (non-zero) constant or you can add a constant to both sides.

So you did the 7/4 correctly, which is 1.75. However, you should find 3^x/5^x instead of 3^x-5^x, you see?

 

[swears]

Yeah, I thought I did that. I had 3^(X)/5^(X) = 7/4

But I turned division to subtraction since I thought that was a 'power rule'

 

[TD]

I see, but that's not correct. You should recheck the properties of powers and logarithms.
You were confusing it with the rule: log(a)-log(b) = log(a/b)

Now, there is a rule: a^x/b^x = (a/b)^x, so you can put the exponent above the entire fraction.

 

[swears]

Oh wow, that's crazy.

So does the exponent get added/multiplied? Or is this rule only for common exponents?

I know i'd have .6^X but I'm not certain what to use for my exponent.

 

[TD]

Look at the rule: it's only for common exponents. Both numerator and denominator had exponent x, which can be put over the entire fraction then.

The next step is to take the logarithm of both sides.

 

[swears]

I got X = -1.0955

Thanks for your help =)

It doesn't look like I can use the same tactic with the 2nd problem though.

 

[TD]

That answer is 'correct' (it's an approximation of the exact answer, when you'd leave the logarithms).

Another rule: x^{y + z} = x^y \cdot x^z

 

[swears]

Aha. I think I got it.

I got x = -3.263

BTW, where can I get one of these 'rule' books?

 

[TD]

Strange, I find something slightly different - did you round correctly?

<br /> 2^x = e^{x + 1} \to x = \frac{1}{{\ln 2 - 1}} \approx - 3.259<br />

Links:
http://oakroadsystems.com/math/expolaws.htm"
http://oakroadsystems.com/math/loglaws.htm"

 

[swears]

Hmm, I think it's close enough. My book says the answer is -3.26. I don't think I am advanced enough to worry about that level of preciseness yet.

I did it differently though. I used more steps. I'm not sure how you did it. It looks like you used another rule to rewrite that.

Thanks for your help again.

 

[TD]

Oh yes, I didn't include any steps - just the exact answer and an approximation.

Good luck