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solve the following: 10^(2x-3) = 0.01
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If \(a^x=b^x\), then \(a=b\). And use \(0.01=\frac1 {100}=\frac1{10^2}=10^{-2}\).
Oops. Really you need is that if \(a^x=a^y\) then \(x=y\).
Use what @klimenkov said above. \[10^{2x - 3} = 0.01\]\[10^{2x - 3} = \frac{1}{100}\]\[10^{2x - 3} = \frac{1}{10^{2}}\]\[10^{2x - 3} = 10^{-2}\]\[2x - 3 = -2\]Solve for x now.
The very first statement is false. Forget it, because if \(2^2=(-2)^2\), but \(2\ne-2\).
What do you mean? I'm not quite following.
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I said that mine statement was false.
Oh ok.
i'm sorry. which statement was false?
would the answer be x = 1/2
yup
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Thanks!
I am so late...but you are correct :)
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