Question

How do you solve for x in `1/log_3x + 1/log _27 x = 4`?

Answer 1

`x=3`

Explanation:

Just a simple thing to note: `1/(log_a(b))=log_b(a)`

Therefore,

`1/(log_3(x))+1/(log_27(x))=4`

`=>log_x(3)+log_x(27)=4`

Using the fact that `log_a(b)+log_a(c)=log_a(bc)`,

`=>log_x(3*27)=4`

`=>log_x(81)=4`

Since `log_a(b)=c` is just `a^c=b,`

`x^4=81` We can solve this equation!

`=>x=root[4] (81)`

`=>x=3` or `x=-3`
For now, you can just ignore the negative answer.

`x=3`

Please ask if you are confused or curious about my answer!

Answer 2

`x=3`

Explanation:

Given equation:

`1/\log_3x+1/\log_{27}x=4`

`1/\log_3x+1/\log_{3^3}x=4`

`1/\log_3x+1/{1/3\log_{3}x}=4\quad (\because \ \log_{a^n}b=1/n\log_a b)`

`1/\log_3x+3/{\log_{3}x}=4`

`4/{\log_{3}x}=4`

`\log_{3}x=4/4`

`\log_3x=1`

`x=3`