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



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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


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treefudg offline Unverified User #
An Unknown Location | 6 years, 1 month ago (1 hour, 23 minutes after post)

try casio5274.54 v27

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Creal Default offline Verified User (6 years, 5 months) Long Term User Shouts: 1 #
Houston, TX, US | 6 years, 1 month ago (4 hours, 8 minutes after post)

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.

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Creal Default offline Verified User (6 years, 5 months) Long Term User Shouts: 1 #
Houston, TX, US | 6 years, 1 month ago (4 hours, 38 minutes after post)

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

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tbuz offline Unverified User #
An Unknown Location | 6 years, 1 month ago (4 hours, 51 minutes after post)

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.

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tekoastro offline Verified User (6 years, 1 month) Long Term User Shouts: 3 #
San Diego, CA, US | 6 years, 1 month ago (3 days, 18 hours after post)

Are you solving for A or for B?

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