How do you solve for x in 1/log_3x + 1/log _27 x = 41log3x+1log27x=4?

2 Answers
Jul 28, 2018

x=3x=3

Explanation:

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

Therefore,

1/(log_3(x))+1/(log_27(x))=41log3(x)+1log27(x)=4

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

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

=>log_x(3*27)=4logx(327)=4

=>log_x(81)=4logx(81)=4

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

x^4=81x4=81 We can solve this equation!

=>x=root[4] (81)x=481

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

x=3x=3

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

x=3x=3

Explanation:

Given equation:

1/\log_3x+1/\log_{27}x=41log3x+1log27x=4

1/\log_3x+1/\log_{3^3}x=41log3x+1log33x=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