Подскажите, , как решить: log2 3=a, log8 18=?

Log8_18 = 1/3 * log2_18 = 1/3(log2_2  + log2_9) =
= 1/3* (1+ 2*log2_3) = 1/3 * (1+2*a) = (1+2a) / 3.
второй

log8_18 = (log2_18) / (log2_8)= (log2_2 + log2_9) / 3 = ( 1 + 2*log2_3) / 3 = =(1+2a)/3