SOLUTION: log2 (3x + 1) + log8 (x - 1) ^ 2 = 6 Show that log8 (x - 1) ^ 2 = 6Hence solve the equation for underline x>0

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Question 1189428: log2 (3x + 1) + log8 (x - 1) ^ 2 = 6
Show that log8 (x - 1) ^ 2 = 6Hence solve the equation for underline x>0

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Unclear...!

What does this mean: "solve the equation for underline x>0"?

Furthermore, the given information appears to be inconsistent, so it is probably shown incorrectly.

Given

log2 (3x + 1) + log8 (x - 1) ^ 2 = 6,

if in fact (as you say we are supposed to show)

log8 (x - 1) ^ 2 = 6

then

log2 (3x + 1) = 0

which makes 3x+1=1, which means x=0; and then log8 (x - 1) ^ 2 is NOT equal to 6.

Re-post, showing the information correctly, and making it clear what the last part of the problem is.


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