# question help: How do I solve this: 79=15*B^(A)+A*2^(A)? - Help.com

This post left anonymously

## How do I solve this: 79=15*B^(A)+A*2^(A)?

I got this question in a Satan-like packet from my math teacher (Happy Easter to her too) and I really need help with answering it. Please help!

This open post was written 6 years, 1 month ago | V/U/S: 334, 5, 5 | Edit Post | Leave a reply | Report Post

### Reciprocity (0)

Since writing this post Anonymous may have helped people, but has not within the last 4 days.

### Replies (5)

Where were you?

Click and drag to move the map around. FAQ: How we place people on this map »
You can also watch events on Help.com as they happen
Mouse over the map for 2 seconds to see an expanded, interactive view

try casio5274.54 v27

This account has been deactivated.

I can’t figure this one out entirely. Seems like there would be an infinite number of solutions for solving for A and B. This is me playing around with the numbers a little:

Take the natural log of both sides and bring down the exponents.
ln(79) = A*ln(15*B) + A*ln(A*2)
Divide both sides by A:
ln(79)/A = ln(15*B) + ln(A*2)
Using a property of logarithms, we can combine the two on the right side:
ln(79)/A = ln(30*A*B)
Multipy by A:
ln(79) = A*ln(30*A*B)
Raise to a power of A and remove the natural logs.
79 = (30*A*B)^A
Sadly, that’s as far as I can simplify it. Good luck.

Actually I don’t think my logic is right on that last post. Nevermind, it’s too tough for me.

You can’t evaluate something with two variables without having an alternate equation defining one variable in terms of the other (e.g. B = A + 6). Then you can substitue ‘A + 6′ for all instances of B and solve in terms of A.

So, what is the final form the teacher is looking for? It cannot be an evaluated expression, so do they just want you to simplify it?

And, what topics are you studying currently? That could help assess the degree of math she wants you to use to solve this.

This account has been deactivated.

Are you solving for A or for B?

### Invite Others to Help

A logged in and verified Help.com member has the ability to setup a Friends List and invite others to help with posts.