## anonymous 5 years ago Solve the equations using properties of logs: a.) 3log[11](x) = 2log[11]8 b.) 2log[5](x) = 3log[5]9

1. anonymous

$3\log_{11}(x)=2\log_{11}8$

2. anonymous

From properties of logs$3\log_{11}x=\log_{11}x^3$and$2\log_{11}8=\log_{11}8^2=\log_{11}64$so the first equation boils down to$\log_{11}x^3=\log_{11}64 \rightarrow x^3=64$so that$x=4$

3. anonymous

$B.) 2\log_{5}(x)=3\log_{5}9$

4. anonymous

It's something similar for the second one too.

5. anonymous

is B.) up there too?

6. anonymous

?

7. anonymous

there were two equations i posted, A and B

8. anonymous

$x^2=9^3=729 \rightarrow x=\pm 27$

9. anonymous

Yeah, I know, but the procedure is the same. The bases are different but it's irrelevant. It's just something to throw you off. As long as the bases are the same in your equation, you can remove the logs in the end using the definition of the log.

10. anonymous

So the answer for B.) is + or - 27?

11. anonymous

yes

12. anonymous

where did the 729 come from?

13. anonymous

9^3?

14. anonymous

729 is 9^3