Question

2 log4+ log4 32- 3log squareroot of 16 2

Answer 1

Answer:

5/2

Step-by-step solution:

(a) indicate "a" is in base.

2log4 + log(4)32 - 3log((16)^1/2)2

simplify

2log4 +  log(4)((4^2)*2) - 3log((2)^3)2

2log4  + 2 + log(4)2 -(3/3)log(2)2

2log4  + 2 + log(2^2)2 -1

2log4  + 2 +(1/2) log(2)2 -1

1  + 2 + 1/2  -1

5/2

Answer 2

Concept:

Logarithms can also be used to express exponents. An integer's base logarithm equals another number. The inverse of exponentiation is a logarithm. Logs having a base of are known as natural logs. Logs having a base of 10 are known as common logs.

Given:

The expression $2log4+log_4 32 -3log_{\sqrt{16}}2$.

Find:

The value of the given expression

Solution:

$2log4+log_4 32 -3log_{\sqrt{16}}2\\= 2log2^2 + log_{2^2}2^5-3log_42\\=4log2+\frac{5}{2}log_22-3log_{2^2}2\\=4log2+\frac{5}{2}-\frac{3}{2}\\=4*0.3+1\\=2.2$

Hence, the value of the expression is $2.2$.