Question

How do you solve this logarithm?

`ln (3x)- ln 2=4`

Answer 1

`x=2/3e^4`

Explanation:

Use the following `color(blue)"laws of logarithms"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(logx-logy=log(x/y))color(white)(2/2)|)))`
This law applies to logarithms in any base.

`rArrln(3x)-ln2=ln((3x)/2)`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(log_b x=nhArrx=b^n)color(white)(2/2)|)))`

Now the natural log of x, written `lnx=log_e x`

`rArrln(3x)-ln2=4`

can be expressed as

`ln_e((3x)/2)=4rArr(3x)/2=e^4`

multiply by 2 and divide by 3

`rArrx=2/3e^4`