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soccergal12

  • 3 years ago

solve the following: 10^(2x-3) = 0.01

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  1. klimenkov
    • 3 years ago
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    If \(a^x=b^x\), then \(a=b\). And use \(0.01=\frac1 {100}=\frac1{10^2}=10^{-2}\).

  2. klimenkov
    • 3 years ago
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    Oops. Really you need is that if \(a^x=a^y\) then \(x=y\).

  3. Calcmathlete
    • 3 years ago
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    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.

  4. klimenkov
    • 3 years ago
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    The very first statement is false. Forget it, because if \(2^2=(-2)^2\), but \(2\ne-2\).

  5. Calcmathlete
    • 3 years ago
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    What do you mean? I'm not quite following.

  6. klimenkov
    • 3 years ago
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    I said that mine statement was false.

  7. Calcmathlete
    • 3 years ago
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    Oh ok.

  8. soccergal12
    • 3 years ago
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    i'm sorry. which statement was false?

  9. soccergal12
    • 3 years ago
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    would the answer be x = 1/2

  10. theyatin
    • 3 years ago
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    yup

  11. soccergal12
    • 3 years ago
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    Thanks!

  12. Calcmathlete
    • 3 years ago
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    I am so late...but you are correct :)

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