dave13 wrote:

Hi

pushpitkc cab you expain based on which rule do we take out \(5^x\) common

i mean if it were like this (2x+7x) i understand we take out common x --> x(2+7)

but in this case \(5^x\) – \(5^{x-3}\) = \((124)\)\((5^y)\) how can we take common \(5^x\) it is not even in the brackets and there is only only one \(5^x\)

Also how after this \(5^x\) * \(\frac{1}{125}\) = \((5^y)\) --- Striking out 124.

we get this \(5^x\) = 5^(y+3)

thanks for your great help

Hey

dave13The first step is to understand what happens to \(5^{x-3}\).

\(5^{x+(-3)} = 5^{x}*5^{-3} = 5^{x}*\frac{1}{5^3} = 5^{x}*\frac{1}{125}\)

Now, \(5^x - 5^{x-3}\) = \(5^x – 5^{x}*\frac{1}{125}\) = \(5^x(1 - \frac{1}{125}) = \frac{124}{125} * 5^x\)

Now, since we have \(124\) on both sides, we will finally get \(5^x = 5^y*125 = 5^y*5^3 = 5^{y+3}\)

Hope this helps you.

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