Solving logarithmic equation, different bases What number do I need to multiply both sides with? I have worked for an hour on this but it is the first time I am using this website so it is impossible for me to write what I have already done. If you can give me tips on how to solve it I would appreciate it a lot! 2 log_6 (sqrt(x) + root[4](x)) = log_4 x

Uroskopieulm

Uroskopieulm

Answered question

2022-11-17

Solving logarithmic equation, different bases
What number do I need to multiply both sides with? I have worked for an hour on this but it is the first time I am using this website so it is impossible for me to write what I have already done. If you can give me tips on how to solve it I would appreciate it a lot!
2 log 6 ( x + x 4 ) = log 4 x

Answer & Explanation

sellk9o

sellk9o

Beginner2022-11-18Added 11 answers

Note that 2 log b u = log b u 2 for any b and u, so that we can rewrite this as
log 6 ( x + 2 x 3 4 + x ) = log 4 x
From here, the easiest thing is to guess, and observe that for the argument of the left-side log comes out nicely if x is a fourth power. For instance, we might guess x = 2 4 = 16, which yields x + 2 x 3 4 + x = 16 + 2 ( 8 ) + 4 = 36, and indeed
log 6 36 = 2 = log 4 16
ETA: That this is the only solution for x > 0 can be seen if we write (following GEdgar's comments to the original post)
log 4 ( x + 2 x 3 4 + x ) log 4 6 = log 4 x
log 4 ( x + 2 x 3 4 + x ) log 4 x = log 4 6
or equivalently
log 4 6 = log x ( x + 2 x 3 4 + x ) = 1 + log x ( 1 + 2 x 4 + 1 x ) = 1 + 2 log x ( 1 + 1 x 4 )
Note that the argument of the log x is a decreasing function of x, which means that the entire right-hand side is a decreasing function of x for x > 1, which in turn means that the solution x = 16 is the only solution for x > 1
For 0 < x < 1, the argument of the log x is greater than 1, but its base x is less than 1, so the log is negative, and the right-hand side is less than 1, whereas the left-hand side is greater than 1, so there are no solutions in that range, either.
Finally, x = 1 is not a solution of the original equation, by inspection.

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