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505-555 (Easy)|   Exponents|            
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Preeeeeetika
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PareshGmat
\(5^x – 5^{(x- 3)} = 124*5^y\)

Re-write as

\(5^x – 5^{(x- 3)} = (125 - 1 ) * 5^y\)

\(5^x – 5^{(x- 3)} = 125* 5^y - 5^y\)

\(5^x – 5^{(x- 3)} = 5^{(y+3)} - 5^y\)

LHS = RHS, so equating the powers

y = x+3 ; Answer = C

Hi PareshGmat,
I don't understand how equating the powers in the last part makes y = x+3. Could you please kindly explain.

Edit: i just noticed. think you wrote it wrong the answer should be y= x-3 not y = x+1
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If we take x = 3, it becomes
\(5^3 – 5^{(3- 3)} = 124*5^y\)
\(5^3 – 5^0 = 124*5^y\)
\(125 – 1 = 124*5^y\)
\(124 = 124*5^y\)
\(5^y = 1\)
y = 0

Substituting x = 3, we get y = 0 for C
alltimeacheiver
If \(5^x – 5^{(x- 3)} = 124*5^y\), what is y in terms of x?

A. x
B. x - 6
C. x - 3
D. 2x + 3
E. 2x + 6
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interesting question - there are two ways to do this

i think one way is safer (i.e., works pretty much every time) and one way works in occasion but more intuitive imo

method 1: factoring (safer but takes a bit longer)
factor out the smaller term on the left hand side (LHS), we have 5^(x-3) * 5^3 = 5^(x-3) * (x^3-1)
we can than identify the equal base term and ensure power is the same, we have 5^(x-3) = 5^y and x^3-1 = 124
y = x-3

method 2: spot unique numbers
some people are better at this and may not work every time but 124 is clear an unique number, i.e., 125-1 and 125 is 5^3
so we can also start with the right hand side (RHS), we have 124 * 5^y = (5^3 - 1) * 5^y = 5^3 * 5^y - 5^y = 5^(3+y) - 5^y
this is much more intuitive and equal to spot the relationship from my pov; we have 5^(x) = 5^(3+y) and 5^(x-3) = y
x = 3+y
y = x-3

the answer is (C) x-3
alltimeacheiver
If \(5^x – 5^{(x- 3)} = 124*5^y\), what is y in terms of x?

A. x
B. x - 6
C. x - 3
D. 2x + 3
E. 2x + 6
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