Question

Expand log_1/2(3x^2/2) using the properties and rules for logarithms.


Please help.

Answer 1

Answer:

$\frac{0.1761+2\log(x)}{-0.3010}$

Step-by-step explanation:

Data provided:

$\log_{1/2}(\frac{3x^2}{2})$

now,

we know the properties of log functions as:

1) log(AB) = log(A) + log(B)

2) $\log(\frac{A}{B})$ = log(A) - log(B)

3) log(xⁿ) = n × log(x)

thus,

$\log_{1/2}(\frac{3x^2}{2})$ = $\log_{1/2}(3x^2)$ - $\log_{1/2}(2)$

or

$\log_{1/2}(\frac{3x^2}{2})$ = $\log_{1/2}(3)+\log_{\frac{1}{2}}(x^2)$ - $\log_{1/2}(2)$

or

using property 3

$\log_{1/2}(\frac{3x^2}{2})$ = $\log_{1/2}(3)+2\log_{\frac{1}{2}}(x)$ - $\log_{1/2}(2)$

also,

$\log_a(x)=\frac{\log(x)}{\log(a)}$

thus,

$\log_{1/2}(\frac{3x^2}{2})$ = $\frac{\log(3)}{\log(\frac{1}{2})}+2\times[\frac{\log(x)}{\log(\frac{1}{2}}]-\frac{\log(2)}{\log(\frac{1}{2})}$

or

$\log_{1/2}(\frac{3x^2}{2})$ = $\frac{\log(3)+2\log(x)-\log(2)}{\log(\frac{1}{2})}$

or

$\log_{1/2}(\frac{3x^2}{2})$ = $\frac{\log(3)+2\log(x)-\log(2)}{\log(1)-log(2)}$

now,

log(1) = 0

log(2) = 0.3010

log(3) = 0.4771

thus,

$\log_{1/2}(\frac{3x^2}{2})$ = $\frac{0.4771+2\log(x)-0.3010}{0-0.3010}$

or

$\log_{1/2}(\frac{3x^2}{2})$ = $\frac{0.1761+2\log(x)}{-0.3010}$

Answer 2

Answer:

-0.58-6.64log(x)

Step-by-step explanation:

The given expression is

$log_{\frac{1}{2}}(\frac{3x^2}{2})$

It can be rewritten as

$log_{\frac{1}{2}}(\frac{1}{2}\times 3\times x^2)$

Using the properties and rules for logarithms, we get

$log_{\frac{1}{2}}(\frac{1}{2})+log_{\frac{1}{2}}(3)+log_{\frac{1}{2}}(x^2)$         $[\because log_a(mn)=log_am+log_an]$

$1+log_{\frac{1}{2}}(3)+2log_{\frac{1}{2}}x$         $[\because log_a(a)=1,logx^n=nlogx]$

$1+\frac{log(3)}{log(\frac{1}{2})}+2\frac{log(x)}{log(\frac{1}{2})}$             $[\because log_a(b)=\frac{logb}{loga}]$

$1+\frac{log(3)}{log(1)-log(2)}+2\frac{log(x)}{log(1)-log(2)}$        $[\because log\frac{a}{b}=loga-logb]$

We know that,

log (1) = 0

log (2) = 0.301

log (3) = 0.477

Substitute these values in the

$1+\frac{0.477}{0-0.301}+2\frac{log(x)}{0-0.301}$

$1-1.58-2\frac{log(x)}{0.301}$

$-0.58-6.64log(x)$

Therefore, the expanded form of given expression is -0.58-6.64log(x).