## anonymous 5 years ago Solving for x... Can anybody help? I think I did it right but want to make sure. Log2 base 2 ^ x-5 = 5

1. anonymous

Do you mean$\log_22^{x-5}$?

2. anonymous

If so, that's$\log_22^{x-5}=(x-5)\log_22=(x-5).1=x-5$

3. anonymous

Yes that is the equation. Solving for X.

4. anonymous

what is X?

5. anonymous

Well, you can't solve for x unless $\log_22^{x-5}=(something)$

6. anonymous

x=10

7. anonymous

8. anonymous

x=10 does that help?

9. anonymous

Sorry - I didn't see the 5

10. anonymous

$x-5=5 \rightarrow x=10$

11. anonymous

Yes It does . Thank you . That is what I got.