Question

2 log base 10 8 +log base 10 36 _ log base 10 (1.5) _ 3 log base 10 2

Answer 1

Given problem is $2 \log_{10}(8) +\log_{10}(36)- \log_{10}(1.5)-3 \log_{10}(2)$

Apply power rule of logarithm which is $m \log_b(x)=\log_b(x^m)$

Using that formula we get:

= $\log_{10}(8^2) +\log_{10}(36)- \log_{10}(1.5)- \log_{10}(2^3)$

= $\log_{10}(64) +\log_{10}(36)- \log_{10}(1.5)- \log_{10}(8)$

Apply product rule of logarithm which is $\log_b(x)+ \log_b(y)=\log_b(x*y)$

Using that formula we get:

= $\log_{10}(64*36)- \log_{10}(1.5)- \log_{10}(8)$

= $\log_{10}(64*36)- (\log_{10}(1.5)+ \log_{10}(8))$

apply product rule again

= $\log_{10}(64*36)- (\log_{10}(1.5*8)$

= $\log_{10}(2304)- (\log_{10}(12)$

Apply quotient rule of logarithm which is $\log_b(x)-\log_b(y)=\log_b(\frac{x}{y})$

Using that formula we get:

= $\log_{10}(\frac{2304}{12})$

= $\log_{10}(192)$

Hence final answer is $\log_{10}(192)$.