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What I'm not sure about is the power of two above the logarithm. I just wanted to verify I'm calculating correctly

Do I do these steps...?

1 - Take absolute value of variable AL

2 - Take log base 10 of result 1

3 - take result 2 squared

4 - take result 3 multiply by variable b2

expression

erotavlas
  • 119

2 Answers2

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Yes, this is correct. $\log^2x$ is short for $(\log x)^2$.

qwr
  • 10,716
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Yes.Your solution is $b_2(\mathrm{\log}_{10 }|AL|)^2$.