(3^m)(5^(m+1))=12^(2m-1)

not sure how to solve this but I think you use natural logs? I'm just stuck because the first two are multiplying and not adding and we didn't go over this

yes, take the log of both sides

log(3^m)(5^(m+1)) = log(12^(2m-1)

mlog3 + (m+1)log5 = (2m-1)log12

expand and you will get terms of m and constants, remember something like log12 is a constant
factor out m from the "m-terms" and solve for m

It is messy and long, be careful.

I just solved for each log individually on my calculator and used decimals to find the answer, and it ended up being right

thanks!

To solve the equation (3^m)(5^(m+1)) = 12^(2m-1), we can actually simplify it without using natural logarithms.

Let's break it down step by step:

First, we can simplify the equation by expanding the exponents:

(3^m)(5^(m+1)) = 12^(2m-1)
(3^m)(5^m)(5^1) = 12^(2m-1)

Next, we can rewrite 12 as a product of prime factors:

12 = 2^2 * 3

We can substitute 12 with its prime factorization:

(3^m)(5^m)(5^1) = (2^2 * 3)^(2m-1)

Now, we can simplify further:

(3^m)(5^m)(5^1) = 2^(2(2m-1)) * 3^(2m-1)

We can distribute the exponent 2(2m-1) to each term in 2^(2m-1):

(3^m)(5^m)(5^1) = 2^(4m-2) * 3^(2m-1)

Since both sides of the equation have the same base 2, we can equate the exponents:

4m-2 = 2m-1

Now we can solve this equation for m:

Subtract 2m from both sides:
4m - 2m - 2 = 2m - 2m - 1
2m - 2 = -1

Add 2 to both sides:
2m - 2 + 2 = -1 + 2
2m = 1

Divide both sides by 2:
2m/2 = 1/2
m = 1/2

So, the value of m that satisfies the equation is 1/2.

There is no need to use natural logarithms in this particular problem. We can solve it by simplifying the equation step by step and using algebraic manipulations.