Given that (2^p)*(8^q)=(2^n), express n in terms of p and q.?

(2^p) * (8^q) = (2^n)
log[(2^p) * (8^q)] = log(2^n)
log(2^p) + log(8^q) = n log(2)
p log(2) + q log(8) = n log(2)
p log(2) + q log(2^3) = n log(2)
p log(2) + 3q log(2) = n log(2)
(p + 3q) log(2) = n log(2)
p + 3q = n
n = p + 3q

First thing to note is that 8 is 2³.

If you have a number that is an exponent of an exponent, it's the same as multiplying the exponents:

(a^x)^y = a^(xy)

So our 8^q becomes (2³)^q = 2^(3q)

So now we have:

2^p * 2^(3q) = 2^n

Now we use this rule: if we multiply two numbers together that have the same base, it's the same as adding the exponents:

a^x * a^y = a^(x y)

So since we're multiplying 2 numbers each with base 2, we add the exponents.

so now we have:

2^(p 3q) = 2^n

And now one more thing. We have two numbers represented as exponents with the same base.

If the value of the bases are the same, the value of the exponents must also be the same. (non-logarithm way of looking at taking the log-base-2 of each side).

So that gives us:

p 3q = n

Which is your answer.

2^(p) * 8^(q) = 2^(n)

2^(p) * (2^3)^(q) = 2^(n)

2^(p) * 2^(3q) = 2^(n)

2^(p + 3q) = 2^(n)

p + 3q = n

n = p + 3q