If log10⁡5=a and log10⁡3=b then

1 log30⁡8=3log10⁡2log10⁡5+log10⁡3+log10⁡2=31−log10⁡5log10⁡5+log10⁡3+log10⁡2=31−log10⁡51+log10⁡3=3(1−a)1+b2 log40⁡15=log10⁡15log10⁡40=log10⁡3+log10⁡5log10⁡5+31−log10⁡5=a+b3−2a (3)  log243⁡32=log10⁡2log10⁡3=1−log10⁡5log10⁡3=1−ab