If log a= x and log10 b= y, express log10 [100a^(3)b^(-1/2)] ÷ in terms of x and y
Using the properties of logarithms:
log10 [100a^(3)b^(-1/2)] = log10 100 + log10 a^(3) - log10 b^(1/2)
= 2 + 3log10 a - (1/2)log10 b
Then, using the given values:
= 2 + 3x - (1/2)y
Therefore, log10 [100a^(3)b^(-1/2)] ÷ = 3x - (1/2)y - 2