Math CentralQuandaries & Queries


Question from Mukulu, a student:

Help me with this question because i tried my level best to get the right answer but i failed
Log_7 (4X) = Log_14 X

Hi Mukulu.

Log7 is how I write "log in base 7". And Log14 is how I write "log in base 14".

So you have Log7 (4x) = Log14 (x).

Step 1: Let y = Log14(x). This gives us:
Log7 (4x) = y.

Step 2: Raise both sides under the base of 7:
7[Log7 (4x)] = 7y
4x = 7y

Step 3: Since y = Log14(x), let's raise this equation under base 14:
14y = 14[Log14(x)]
14y = x.

Step 4: Substitute x from Step 3 into Step 2 equations:
4 [ 14y] = 7y
4 (2y)(7y) = 7y
4 (2y) = 1

Can you finish the problem now? x works out to a nice simple number.

Stephen La Rocque.

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