Question

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


Please help.

Answer 1

`(0.1761+2\log(x))/(-0.3010)`

Explanation:

Data provided:

`\log_(1/2)((3x^2)/(2))`

now,

we know the properties of log functions as:

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

2)
`\log((A)/(B))` = log(A) - log(B)

3) log(xⁿ) = n × log(x)

thus,

`\log_(1/2)((3x^2)/(2))` =
`\log_(1/2)(3x^2)` -
`\log_(1/2)(2)`

or

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

or

using property 3

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

also,

`\log_a(x)=(\log(x))/(\log(a))`

thus,

`\log_(1/2)((3x^2)/(2))` =
`(\log(3))/(\log((1)/(2)))+2*[(\log(x))/(\log((1)/(2))]-(\log(2))/(\log((1)/(2)))`

or

`\log_(1/2)((3x^2)/(2))` =
`(\log(3)+2\log(x)-\log(2))/(\log((1)/(2)))`

or

`\log_(1/2)((3x^2)/(2))` =
`(\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)((3x^2)/(2))` =
`(0.4771+2\log(x)-0.3010)/(0-0.3010)`

or

`\log_(1/2)((3x^2)/(2))` =
`(0.1761+2\log(x))/(-0.3010)`

Answer 2

-0.58-6.64log(x)

Explanation:

The given expression is

`log_{(1)/(2)}((3x^2)/(2))`

It can be rewritten as

`log_{(1)/(2)}((1)/(2)* 3* x^2)`

Using the properties and rules for logarithms, we get

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

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

`1+(log(3))/(log((1)/(2)))+2(log(x))/(log((1)/(2)))`
`[\because log_a(b)=(logb)/(loga)]`

`1+(log(3))/(log(1)-log(2))+2(log(x))/(log(1)-log(2))`
`[\because log(a)/(b)=loga-logb]`

We know that,

log (1) = 0

log (2) = 0.301

log (3) = 0.477

Substitute these values in the

`1+(0.477)/(0-0.301)+2(log(x))/(0-0.301)`

`1-1.58-2(log(x))/(0.301)`

`-0.58-6.64log(x)`

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