If 5^x – 5^(x- 3) = 124*5^y, what is y in terms of x?

Finding one variable in terms of another is very simple. Assume a value for one and find the other. Then look for the option that works.

Say, x = 3 (so that x-3 becomes 0 and we are left with simple calculations)

\(5^3 - 5^{3- 3} = (124)(5^y)\)
\(1 = 5^y\)
y = 0

Now put x = 3 in the options and find out which option gives you y = 0. Only C does so answer is (C)

One option is to rewrite the left side of the equation by factoring out 5^(x-3)
So, we get: 5^(x-3)[5^3 - 1] = (124)(5^y)
Evaluate to get: 5^(x-3)[125 - 1] = (124)(5^y)
Simplify to get: 5^(x-3)[124] = (124)(5^y)
Divide both sides by 124 to get: 5^(x-3) = 5^y
So, x-3 = y

Answer: C

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ASIDE: A lot of students struggle to see how we can factor 5^x - 5^(x-3) to get 5^(x-3)[5^3 - 1]
Sure, they may be okay with straightforward factoring like these examples:
k^5 - k^3 = k^3(k^2 - 1)
m^19 - m^15 = m^15(m^4 - 1)
But they have problems when the exponents are variables.

IMPORTANT: Notice that, each time, the greatest common factor of both terms is the term with the smaller exponent.

So, in the expression 5^x - 5^(x-3), the term with the smaller exponent is 5^(x-3, so we can factor out 5^(x-3)

Likewise, w^x + x^(x+5) = w^x(1 + w^5)
And 2^x - 2^(x-2) = 2^(x-2)[2^2 - 1]
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Cheers,
Brent

Another option is to PLUG IN a value for x and see what kind of relationship we get between x and y.
There are two "nice" x-values to plug in. They are x = 0 and x = 3, since we can easily use these to evaluate 5^x and 5^(x-3). Of these two values, x = 3 is the easier one to plug in.
So, let's plug in x = 3
We get: 5^3 - 5^(3-3) = (124)(5^y)
Simplify to get: 125 - 1 = (124)(5^y)
Simplify to get: 124 = (124)(5^y)
Divide both sides by 124 to get: 1 = 5^y
Solve for y to get: y = 0

So, when x = 3, y = 0.

Now we'll check the answer choices to see which one satisfies this relationship.
A) y = x... So, we get 0 = 3 (NOPE)
B) y = x - 6... So, we get 0 = 3 - 6 (NOPE)
C) y = x - 3... So, we get 0 = 3 - 3 IT WORKS!
D) y = 2x + 3... So, we get 0 = 2(3) + 3 (NOPE)
E) y = 2x + 6... So, we get 0 = 2(3) + 6 (NOPE)

Answer: C

Cheers,
Brent

u can also plug in x=3. makes it very simple.

Hi Bunuel,
I', trying to understand your factorization, but I find some troubles when I have to factorize some value with exponent an expression like in the question (x-3 or x+4),
Could you explain me that or do you have a deck where I could dive into this argument?
Thank you in advance

\(5^{x-3}(5^3-1)=124*5^y\) --> \(5^{x-3+3}-5^{x-3}=124*5^y\) --> \(5^{x}-5^{x-3}=124*5^y\).

Theory: math-number-theory-88376.html

Practice:
tough-and-tricky-exponents-and-roots-questions-125956.html
tough-and-tricky-exponents-and-roots-questions-125967.html

Hope it helps.

\({5^{x}}\frac{124}{125}=124(5^y)\)
\(5^{x}=5^{y+3}\)
\(x=y+3-->y=x-3\)

Answer: C

Given that 5^x – 5^(x- 3) = 124*5^y
So, 5^(x-3+3) - 5^(x- 3) = 124*5^y
So, 5^(x-3) [5^3 - 1] = 124*5^y
So, 5^(x-3)*124 = 124*5^y
So, 5^(x-3) = 5^y
So, x-3 = y

Hence option (C).

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ok this is how you solve this

you have 5^x -5^x-3 = (124) (5^y)
step 1: 5^x (1-5^-3) =(124) (5^y)
Step 2: 5^x(1 - 1/5^3) = (124) (5^y)
Step 3: 5^x(1- 1/125) = (124) (5^y)
Step 4: 5^x (125-1)/125 = (124) (5^y)
Step 5: 5^x (124/125) = (124) (5^y)
Step 6: 5^x =[(124) (5^y) (125) ]/124
Step 7: 5^x = (125) (5^y)
Step 8: 5^x = 5^3 (5^y)
Step 9: x = 3 +y --> y = x -3
for explaining purposes I broke it down in to 9 steps. It should not take you more than 5 steps to solve this in term of y.

\(5^x - 5^{x-3} = (124)(5^y)\)

i.e. \(5^{x-3}(5^3 - 1) = (124)(5^y)\)

i.e. \(5^{x-3}(124) = (124)(5^y)\)

i.e. \(5^{x-3} = (5^y)\)

i.e. y = x-3

Answer: Option C

We can simplify the given equation:

5^x – 5^(x- 3) = 124*5^y

5^x – (5^x)(5^- 3) = 124*5^y

5^x(1 – (5^- 3)) = 124*5^y

5^x(1 – 1/125) = 124*5^y

5^x(1 – 1/125) = 124*5^y

5^x(124/125) = 124*5^y

5^x = 124*5^y*(125/124)

5^x = 5^y(125)

5^x = 5^y(5^3)

5^x = 5^y+3

x = y + 3

x - 3 = y

Answer: C

\(5^x\) – 5^(x–3) = (124)\((5^y)\)
or \(5^x\) – \(\frac{5^x}{5^3}\) = \((125-1)\)\((5^y)\)
or \(\frac{(5^x*5^3-5^x)}{5^3}\) = \((5^3-1)\)\((5^y)\)
or \(5^x*(5^3-1)\) = \((5^3-1)\)\((5^y)(5^3)\)
or \(5^x\) = 5^(y+3)
==> \(x=y+3\)
==>\(y=x-3\) ..................Thus the answer is option C.

5^x - 5^(x - 3) = (124)(5^y)

We see that 5^(x - 3) is common to both terms on the left side of the equation, so we factor it out:

5^(x - 3)(5^3 - 1) = (124)(5^y)

5^(x - 3)(124) = (124)(5^y)

5^(x - 3) = 5^y

Recall that when the bases are equal, we can equate the exponents:

x - 3 = y

Answer: C

Could you help me understand the specific steps you took to get from the first bold line to the second? I'm not sure how you managed to get rid of the fraction here. Thanks!

Response:

Notice that in the first bold line, the fraction 124/125 is not in the exponent; that fraction is simply being multiplied by 5^x (as a matter of fact, nothing besides x is in the exponent on and after the third line). Thus, to get rid of the fraction 124/125 on the left hand side, I simply multiplied each side of the equation by 125/124.

travelwithbrandon
It's mentioned in post that it is a sub-600 level question however I find it better than sub 600 level.

Its mentioned in the left hand corner just before the question starts. sub 600