OpenStudy (anonymous):

rearrange the equation to isolate A. H=K+log(A/C)

5 years ago
OpenStudy (anonymous):

This is what I've done and I keep getting it wrong. These are the steps the hint provides. Subrtract k, then take the antilog of both sides, and then multiply both sides by C. \[H-K=\log{\frac{ A }{ C }} \] \[10^{H-K}=\frac{ A }{C }\] I get: \[10C ^{H-K}=A\]

5 years ago
OpenStudy (whpalmer4):

Let's pretend that H = 5, and K = 3. Furthermore, C = 5. What is \(C*10^{H-K}\)?

5 years ago
OpenStudy (whpalmer4):

@Dama28 still there?

5 years ago
OpenStudy (anonymous):

yeah sorry

5 years ago
OpenStudy (whpalmer4):

So, you've had a little time now, what is the value of \(5*10^{(5-3)} = 5*10^2\) Does that suggest what your mistake was?

5 years ago
OpenStudy (anonymous):

so it should be C*10(H-K)=a??

5 years ago
OpenStudy (anonymous):

sorry C*10^(H-K)=A

5 years ago
OpenStudy (whpalmer4):

yes!

5 years ago
OpenStudy (anonymous):

omg! thank you so much! hahahah I would have never thought of that detail.

5 years ago