Question

How do you solve `2log_2x-log_2 5=2^2`?

Answer 1

`x=4sqrt5`

Explanation:

As `log_ba=loga/logb`

`2log_2 x-log_2 5=2^2` can be written as

`2logx/log2-log5/log2=4` and

multiplying each side by `log2` as `log2!=0`, we get

`2logx-log5=4log2` or

`x^2/5=2^4` or `x^2=80` or

`x=sqrt80=4sqrt5` (as `x` cannot be negative)

Answer 2

`color(blue)(=>x=4sqrt(5))`

Explanation:

Example: I have chosen to use log to base 10 so that you can check it on your calculator if you so wish.

suppose we had `Log_10(z)=2`

This means`-> 10^2=z`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the value of "x)`

Given: `" "2log_2x-log_2 5=2^2`

Write as: `log_2(x^2) -log_2( 5)=4`

Subtraction of logs means that the source value are applying division

`=> log_2(x^2/5)=4`

Using the principle demonstrated in my example

`=> log_2(x^2/5)=4" "->" "2^4=x^2/5`

`=> x=sqrt(80)" "=" " sqrt( 2^2xx2^2xx5)`

`color(blue)(=>x=4sqrt(5))`