can you help me out with this math problem? Show that [x^a(b-c)]/[x^b(a-c)] divided by [(x^b/x^a)^c]=1?

Question

please do not solve this using logarithm method; my teacher told me specifically only to solve this arithmetically

Answer 1

x^a(b-c) = x^(ab - ac) = x^(ab)/x^(ac)

x^b(a-c) = x^(ab - bc) = x^(ab)/x^(bc)

Now, [x^a(b-c)]/[x^b(a-c)] =
= [x^(ab)/x^(ac)] / [x^(ab)/x^(bc)] =
= [1/x^(ac)]/[1/x^(bc)] =
= 1/x^(ac) * x^(bc) =
= x^(bc)/x^(ac) =
= [(x^b)^c]/[(x^a)^c] =
= [(x^b/x^a)^c]

QED

Answer 2

Let's start with the first half of the problem:
[x^a(b-c)]/[x^b(a-c)]

You can bring the denominator to the top by making the exponent negative:
[x^a(b-c)] * {x^-[b(a-c)]}

When you do all the distributions, you get:
(x^ab-ac) * (x^-ba+bc) or
(x^ab-ac) * (x^-ab+bc)

When you multiply two (same) bases that have exponents, you simply add the exponents, keeping the base:
x^bc-ac

(The ab terms got canceled)

Now we can apply the same rules to the second part of the problem:
[(x^b/x^a)^c]

Remember that this whole term was in the denominator, so we can bring c to the top by reversing its sign:
[(x^b/x^a)^-c]

And we can do the same for a:
[(x^b * x^-a)^-c]=

Now when you have a power raised to another power, you simply multiply the exponents:
(x^-bc * x^ac)

Adding these exponents, we get:
(x^-bc+ac)

Now our full problem looks like:
x^(bc-ac) * x^(-bc+ac) = 1

Now when you add those exponents, you get zero:
x^0 = 1

And anything raised to the zero power is 1:
1 = 1

Answer 3

x^[a(b-c)]/[x^b(a-c)]=x^[ab-ac-ab+bc]=x^(bc-ac)=x^c(b-a).
=[x^b/x^a)^c].
this easily leads to the final result desired.

Answer 4

[x^a(b-c)]/[x^b(a-c)] divided by [(x^b/x^a)^c]
[x^ab-ac /x^ab-bc ] divided by [(x^b-a)^c]
x^ab-ac -ab +bc divided by x^c(b-a)
x^c(b-a) divided by x^c(b-a)
=1