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Help with simple logarithms?
I need help with a few questions!! I honestly don't know how to solve them.
1. 7^x = 21
2. 3^x = 12
Thank you for the help!! Please also give a brief explanation how to do it!
4 Answers
- Anonymous9 years agoFavorite Answer
7^x = 21
take the log of each side and apply the power rule...
log (7^x) = x log 7 = log 21
x = (log 21)/(log 7)
from your calculator...x ≈ 1.5646
you try the second one...try !
id est
[x ≈ 2.2619]
- 9 years ago
Logarithms are a way of finding powers. Numbers to powers are bases, where 3^2 has a base of 3. This is why the log of 21 in base 1.7 gives you x. However, sometimes you have to calculate things in base 10, which require taking the log of both sides. So, either log(b1.7)21=x OR log1.7^x=log21 (all in base 10)
log1.7^x=log21
xlog1.7=log21
x=log21/log1.7
The same principle applies for 2.
- 9 years ago
For the first one, do log21/log7 which equals 1.56458
For the second one do log12/log3 which equals 2.26189
- pameliaLv 45 years ago
truly uncomplicated ones. a): 2/3 logx + a million/3 logxy^3=2/3 logx+a million/3(logx+logy^3)=2/3 logx+a million/3logx+a million/3logy^3=logx+logy=log(xy) b):log (x^2 + 5x) - log (x+5)=log(x(x+5))-log(x+5)= logx+log(x+5)-log(x+5)=logx c):log(x^2-sixty 4)-log(x+8)=log[(x+8)(x-8)] -log(x+8)= log(x+8)+log(x-8)-log(x+8)=log(x-8)
