Question

hello! So the people in Indian server were bad(as they would not answer and instead spam) so I switched servers and I hope you all will help me! pleaase​

Answer 1

Answer:

$\displaystyle E=26log2-24log3+8log 5$

Step-by-step explanation:

Logarithms

There are certain properties of the logarithms we'll need to recall in order to simplify the expression given in the question. We'll list them below

$log(x.y)=logx+logy\\log(x/y)=logx-logy\\logx^y=y.logx$

We'll also need to factor the following numbers into their prime factors:

$20=2^2\cdot 5$

$16=2^4$

$15=3\cdot 5$

$25=5^2$

$24=2^3\cdot 3$

$81=3^4$

$80=2^4\cdot 5$

The original expression is

$\displaystyle E=log20+16log\frac{16}{15}+12log\frac{25}{24}+log\frac{81}{80}$

Plugging in the factored values

$\displaystyle E=log(2^2\cdot 5)+16log\frac{2^4}{3\cdot 5}+12log\frac{5^2}{2^3\cdot 3}+log\frac{3^4}{2^4\cdot 5}$

Applying the properties we have

$\displaystyle E=log2^2+log 5+16(log2^4-log3-log 5)+12(log5^2-log2^3-log 3)+log3^4-log2^4-log 5$

$\displaystyle E=2log2+log 5+16log2^4-16log3-16log 5+12log5^2-12log2^3-12log 3+log3^4-log2^4-log 5$

$\displaystyle E=2log2+log 5+64log2-16log3-16log 5+24log5-36log2-12log 3+4log3-4log2-log 5$

Simplifying

$\boxed{\displaystyle E=26log2-24log3+8log 5}$