Question

valor de S=log1/2 32+ log10 (0.001)- log0.1(10√¯10)

Answer 1

$<var>S=A+B-C\\\\temos:\\A=log_{\frac{1}{2}}32=>(2^{-1})^x=2^5=> x=-5\\\\B=log(0,001)=>10^x=\frac{1}{1000}=>10^x=10^{-3}=>x=-3\\\\C=log_{(0,1)}(10^\sqrt{10})=>log_{\frac{1}{10}}(10^1.10^{\frac{1}{2}})=>(10^{-1})^x=10^{\frac{3}{2}}=>x=-\frac{3}{2}}\\\\Substituindo\\S=A+B-C\\S=(-5)+(-3)-(-\frac{3}{2})=>-8+\frac{3}{2}=>S=-\frac{13}{2} </var>$