I don’t think there is. It has been given as a question but I always understood that unlike exponents couldn’t be added.
I can re-write it as 3^2m + 3^(3-2m) = 28 but I’m not sure how that would help me.
Any help would be most appreciated.
Cheers,
Adrian

Actually, it is solvable. Start by noting that:

3^{(3-2m)} =\frac {3^3}{3^{2m}}=\frac {27}{9^m}

Then multiply the equation by 9^m. This gives an equation with terms having factors of 9^m and 9^{2m}. At this point you can substitute a new variable for 9^{m}; for example let x=9^{m}. Now you have a quadratic equation in x, which you can factor to solve for x. Finally convert back using \log 9^m = log x, so:

m = \frac {\log x}{\log 9}

I suggest you use log base 3 for that. Good luck, and post back with your solutions (there are two).

Wow! Okay, so I got to (9^m-1)(9^m-27)=0 and so the solutions are m = 0 or m= 3/2. Correct? When I substituted the values, they both worked in the original equation. Thanks so much.

the solutions are m = 0 or m= 3/2. Correct?

Yes - correct!

I just noticed that the question said = 28. Not 27.

I just noticed that the question said = 28. Not 27

Right - that's why it factors the way that Adrian showed. If the problem had been =27 the solution would have been a bit messier.

Been too long since I did this kind of math. Oy. Just looked off.

Okay, I’m afraid I’ve hit upon another stumbling block.
The original question is: log10(8x) - log10(1+sqrtx) = 2
I got it to 10^2 = 8x/(1 +sqrtx)
Now I get confused.
If I multiply both sides by (1 + sqrtx) I get
100 + 100sqrtx = 8x
Can you let me know if I’m on the right track?
Thanks.

Yes, you're on the right track. Substitute w = sqrt(x) and see where it takes you. Warning: it's a bit of a mess.

Oh gosh, that’s a step I forgot about. So substituting I get 100 + 100w = 8w^2
Using the quadratic formula, I got [100 +/- sqrt(100^2 + 320)]/16. So the result for w was 13.43 or -0.93. Substituting back and squaring, I finally arrived at x = 180.37 or x = 0.87. How did I do?

I finally arrived at x = 180.37 or x = 0.87. How did I do?

Pretty good, except... try substituting those values back into the original equation and see if they work. One does, but the other does not.

Right. I got w = -0.93 & 13.43 and when substituted back as sqrtx, the former is not possible, so there’s only one answer which is (13.43)^2 or x = 180.37. Really appreciate your input. Thanks a lot.

Okay. Hi and thanks again for the help.
Needless to say I have one more question that needs a bit of input.
The population of New York and LA aregrowing at 1% and 1.4% a year. Starting from 8 million (NY) and 6 million(LA), when will they be equal?
My reasoning is that this works similar to compound interest, so
6,000,000 x (1.014^n) = 8,000,000 x (1.01^n)
This can be simplified to
0.75 x 1.014^n = 1.01^n
Now I’m stuck again. I’d appreciate a pointer or two but please don’t solve it for me. How can I deal with unlike bases with the same indices?
Cheers. Adrian

Take the log of both sides, and simplify to get n by itself.

Okay, well, honestly that’s confusing. Or should I say asks more questions than it solves, ha!
1). Why would I use logarithms in a situation like this?
2). Can it be solved without the use of logarithms?
3). My limited understanding of logarithms is using the basic formula of a^x = y is the same as loga(y) = x. How am I then allowed to ‘log’ both sides? Surely, if I substitute this formula into my original equation, I would get x = 1.01^n? And that seems to complicate matters.
Sorry to sound like a numbskull but I want to understand the link before I just plough ahead and arrive at an answer without knowing why.
Cheers, Adrian

Okay, well, honestly that's confusing. Or should I say asks more questions than it solves, ha!
1). Why would I use logarithms in a situation like this?

Logarithms are a usefull tool for working problems involving exponents.


Can it be solved without the use of logarithms?

The only other technique i can think of is to use a numerical technique - essentuially making a series of guesses as to the answer and zeroing in until the error is "acceptable," although you may not get an exact solution this way.


My limited understanding of logarithms is using the basic formula of a^x = y is the same as loga(y) = x. How am I then allowed to 'log' both sides?

You can start with:

\log (0.75(1.014^n)) = \log (1.01^n)

Recall that \log (ab) = \log a + \log b and \log a^b = b \log a; hence the equation can be rewritten as:

\log (0.75) + n \log (1.014)= n \log (1.01)

Can you finish it up from here?

Yep, that’s a real big help. Thanks a bundle.
Working out the logs, then subtracting the ’n’ values, then dividing by log(0.75), I got 72.75 years which checks when I replace the values in the original equation.
Really appreciate your input.
Cheers, Adrian