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Can anyone check my answer please?
- Thread starter Soso
- Start date
I'm confused as to what log3 50 means. Does it mean (log3) 50, 50 times the log of 3? Does it mean log(3 (50)), the log of three time 50? Does it mean log(3 50), the log of three hundred and fifty? ...find the integer N such that: N < log3 50 < N+1
I got
1<log3 50 <4
1 < log3 50 < 3+1
N= 1,4
Is that correct?
I'm confused as to what log3 50 means. Does it mean (log3) 50, 50 times the log of 3? Does it mean log(3 (50)), the log of three time 50? Does it mean log(3 50), the log of three hundred and fifty? ...
I do not how can I write it correctly, but 3 is the exponent which usually comes under the letter g of the word log. Hopefully it is clear now
No.
Show your work...
I have already showed my work
Last edited:
Do you mean that log3 50 represent the log of 50 base 3? If so, the equation for transforming bases isI do not how can I write it correctly, but 3 is the exponent which usually comes under the letter g of the word log. Hopefully it is clear now
\(\displaystyle log_b(x) =\, \dfrac{log_c(x)}{log_c(b)}\, =\, \dfrac{ln(x)}{ln(b)}\)
Do you mean that log3 50 represent the log of 50 base 3? If so, the equation for transforming bases is
\(\displaystyle log_b(x) =\, \dfrac{log_c(x)}{log_c(b)}\, =\, \dfrac{ln(x)}{ln(b)}\)
yes I mean like what you said but I think I do not have to use this formula in my question
log(base3) of 50 = log(50)/log(3) = 3.560876...
OK?
the question is to find the value of N
3 < 50 < 81
3 < 50 < 3^4
log(base3) < log(base3) 50 <log(base3)3^4
1 < log (base3) 50 < 4
1 < log (base3) 50 <3+1
log(50)/log(3) = 3.561
this is my answer
N < log3 50 < N+1
3 < 3.561 < 4
So N = 3
yes. So, my first answer is true. I do not know why you said NO when I once asked whether my answer is correct.
Anyway, thank you for your help
Last edited:
Hello, Soso!
\(\displaystyle \begin{array}{cccccc} 27 &<& 50 &<& 81 \\ \\
\log_327 & < & \log_350 &<& \log_381 \\ \\
3 & < & \log_350 &< & 4 \end{array}\)
Therefore: .\(\displaystyle \;N\,=\, 3\)
Find the integer \(\displaystyle N\) such that: . \(\displaystyle N \:<\: \log_3 50 \:<\: N+1\)
\(\displaystyle \begin{array}{cccccc} 27 &<& 50 &<& 81 \\ \\
\log_327 & < & \log_350 &<& \log_381 \\ \\
3 & < & \log_350 &< & 4 \end{array}\)
Therefore: .\(\displaystyle \;N\,=\, 3\)