## x3_drummerchick one year ago how do i solve this? log (1/sqrt1000) will give medals!

1. x3_drummerchick

|dw:1442612409101:dw|

2. zzr0ck3r

$$\log_a(b)=x$$ means $$a^x=b$$ You have $$\log_{10}(\dfrac{1}{1000^{\frac{1}{2}}})=\log_{10}1000^{-\frac{1}{2}}=\log_{10}((10^3)^{-\frac{1}{2}})=\log_{10}10^{-1.5}$$ So $$10^?=10^{-1.5}$$?

3. x3_drummerchick

-1.5?