## anonymous one year ago Y=3-logx How is this done??

1. UnkleRhaukus

Is that $y = \log_3(x)$? use the log identity $y = \log_b(x)\qquad\iff\qquad b^y=x$

2. anonymous

Yeah I know that but how is it done with a negative between 3 and logx

3. anonymous

the answer is y=10^3-y but how

4. UnkleRhaukus

oh it is a negative? like this $y = 3-\log(x)$?

5. UnkleRhaukus

sorry for my earlier misinterpretation.

6. anonymous

No its suppose to be a negative between the three and y

7. UnkleRhaukus

oh , right i erred $y = 3 - \log (x)\\ \log(x) = 3-y\\ x = 10^{3-y}$

8. UnkleRhaukus

Is this what you were looking for?

9. anonymous

Wait so what exactly did you do?

10. UnkleRhaukus

$y = 3 - \log (x)$add log x to both sides of the equation, $y+\log(x) = 3$take away y from both sides .$\log(x) = 3-y$then applied log identity $x = 10^{3-y}$

11. anonymous

No I think I got it now! Thank you!