Question

please try to find the

Answer 1

2log3+3log5-5log2
=log3^2 +log5^3 -log2^5
=log9 +log125 -log32
=log(9×125/32)
=log1125/32
=log35.15

Answer 2

Given Equation is 2log3 + 3log5 - 5log2.


We know that alog(b) = log(b^a).

= log(3^2) + log(5^3) - 5log2

We know that log a + log b = log ab.

= log(3^2 * 5^2) - 5 log 2

= log(9 * 125) - log(2^5)

= log(1125) - log(2^5)

= log(1125) - log(2^5)

= log(1125) - log(32)

It is in the form of log a- log b = log ab.

log(1125/log32) to the base 10.


Therefore the required equation will be log10(1125/32).


Hope this helps!