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The solution set of equation
mod(1-log1/6x)+2=mod(3-log1/6x) is?

Mostafijur Rahaman , 10 Years ago
Grade 12
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

|1-log_{1/6}x| + 2 = |3 - log_{1/6}x|
|log_{1/6}x - 1| + 2 = |log_{1/6}x - 3|
Domain: x > 0

0 < x < 1/216
-log_{1/6}x + 1 + 2 = -log_{1/6}x + 3
3=3
\frac{1}{216}\leq x\leq \frac{1}{6}
-log_{1/6}x + 1 + 2 = log_{1/6}x - 3
2log_{1/6}x = 6
log_{1/6}x = 3
x = \frac{1}{216}
x > 1/6
log_{1/6}x - 1 + 2 = log_{1/6}x - 3
So x should be less than 1/6.


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