Question

Log (1/320) base to 2+1/log 2 base to 10+5

Answer 1

Answer:

$log_{2}( \frac{1}{320} ) + \frac{1}{ log_{10}(2) } + 5 =$

$log_{2}( \frac{1}{320} ) + log_{2}(10) + 5 =$

$log_{2}( \frac{1}{320} \times 10) + 5 =$

$log_{2}( \frac{1}{32} ) + 5 =$

$log_{2}( 2 ) {}^{ - 5} + 5 = - 5 + 5 = 0 \: ans.$

Answer 2

Answer:

same as above solution. nice work