15^x + 15^(x+1) = 60^yajmoney09 wrote:Need Some help please! OA: A
15^x + 15^x * 15 = 15^y * 4^y
factor 15^x out:
15^x(1+15) = 15^y * 4^y
15^x (16) = 15^y *4^y
15^x = 15^y
16 = 4^y
since 16 = 4^2, then y = 2
15^x = 15^y
x = 2
Solving For X
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15^x + 15^x * 15 = 15^y * 4^y
factor 15^x out:
15^x(1+15) = 15^y * 4^y
15^x (16) = 15^y *4^y
15^x = 15^y
16 = 4^y
since 16 = 4^2, then y = 2
15^x = 15^y
x = 2
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Tani
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15^x = 15^y only if x=y. You are not given that and therefore cannot make that assumption.
However, you can manipulate the equation to get (15^x)/(15^y) = (4^y)/(4^2) or 15^(x-y) = 4^(y-2). Looking at this equation we see an integer power of an odd number equal to an integer power of an even number. The only way that is possible is if both are raised to the power of zero. Therefore both x-y and y-2 must be zero and x and y must both equal 2.
However, you can manipulate the equation to get (15^x)/(15^y) = (4^y)/(4^2) or 15^(x-y) = 4^(y-2). Looking at this equation we see an integer power of an odd number equal to an integer power of an even number. The only way that is possible is if both are raised to the power of zero. Therefore both x-y and y-2 must be zero and x and y must both equal 2.
Tani Wolff
Thanks for your help but not sure where the 4^2 is coming from?Tani Wolff - Kaplan wrote:15^x = 15^y only if x=y. You are not given that and therefore cannot make that assumption.
However, you can manipulate the equation to get (15^x)/(15^y) = (4^y)/(4^2) or 15^(x-y) = 4^(y-2). Looking at this equation we see an integer power of an odd number equal to an integer power of an even number. The only way that is possible is if both are raised to the power of zero. Therefore both x-y and y-2 must be zero and x and y must both equal 2.
