Logarithmic function with strange bases Given log_(4n) 40 sqrt(3) = log_(3n) 45, find n. I have rewritten log_(3n) 45 as (log_(4n)45)/(log_(4n)3n) and multiplied to get log_(4n) 40sqrt(3)* log_(4n)3n = log_(4n) 45

John Landry

John Landry

Answered question

2022-07-21

Logarithmic function with strange bases
Given log 4 n 40 3 = log 3 n 45, find n.
I have rewritten log 3 n 45 as log 4 n 45 log 4 n 3 n and multiplied to get
log 4 n 40 3 log 4 n 3 n = log 4 n 45
but do not know how to continue. Hints would be greatly appreciated, but please don't give me the answer. Thank you.

Answer & Explanation

Monica Dennis

Monica Dennis

Beginner2022-07-22Added 13 answers

Since log a b = ln b ln a , where ln is the natural logarithm. So log 4 n 40 3 = log 3 n 45 can be written as
log 4 n 40 3 = ln 40 3 ln 4 n = log 3 n 45 = ln 45 ln 3 n
. Then
ln 40 3 ln n + ln 4 = ln 45 ln n + ln 3 .
Hence we can easily get ln n and then get n.
equissupnica7

equissupnica7

Beginner2022-07-23Added 4 answers

Hint:
log 40 3 log 4 n = log 4 n 40 3 = log 3 n 45 = log 45 log 3 n

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