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Mathematics 18 Online
OpenStudy (anonymous):

Solve: (1/10)^y=100^(y+3)

OpenStudy (whpalmer4):

\[(\frac{1}{10})^y = 100^{(y+3)}\] Some useful properties of exponents: \[(\frac{a}{b})^n = \frac{a^n}{b^n}\] \[a^{-n} = \frac{1}{a^n}\]

OpenStudy (whpalmer4):

and \[(ab)^n = a^nb^n\]

OpenStudy (anonymous):

just what I was about to add :)

OpenStudy (whpalmer4):

So, I would turn the expression on the left into something more like the expression on the right via the 2nd property

OpenStudy (anonymous):

\[1^{y}/10y \] = .... how do you solve\[100 ^{y+3}\] with maybe laws of exponents? I'm not sure

OpenStudy (anonymous):

****[10^{y}\]

OpenStudy (anonymous):

please help:)

OpenStudy (whpalmer4):

Can't you write \[100^{y+3} = 10^{y+3}10^{y+3}\]?

OpenStudy (anonymous):

like \[(10^{2})^{y+3}\]?

OpenStudy (anonymous):

\[10^{-y}=100^{y+3}=10^{10y+30}\] thus \[-y= 10y+30\]

OpenStudy (anonymous):

heheh i mean -y=2y+6

OpenStudy (whpalmer4):

Okay, here's how I would do it: \[y^{-1} = 100^{y+3}\]split rhs \[y^{-1}=10^{y+3}10^{y+3}\]Multiply both sides by \(10^y\) \[1= 10^0 =10^y10^{y+3}10^{y+3}\]Log base 10 of both sides \[0= y+y+3+y+3\]\[y=-2\]

OpenStudy (whpalmer4):

sorry, miswrote the left hand side in the first two lines, should have been \(10^{-y}\)

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