View Full Version : I'm stuck.I hate logarithms
Ashley816
Apr 12, 2007, 02:05 PM
After log the # after is down low and the # after ^ is the exponent a^2 is a squared. Sorry its so confusing...
1. log5^9
2. log5^3/2
3.log5^30
4.log5^10/9
5.log4^a+log4^9=log4^27
6.log7^24-log7^(y+5)=log7^8
7. 2log10^6-1/3log10^27=log10^x
8. log6^(a^2+2)+log6^2=2
galactus
Apr 12, 2007, 02:19 PM
After log the # after is down low and the # after ^ is the exponent a^2 is a squared. Sorry its so confusing...
1. \;\ log_{5}9
2. \;\ log_{5}\frac{3}{2}
3. \;\ log_{5}30
4. \;\ log_{5}\frac{10}{9}
5. \;\ log_{4}a+log_{4}9=log_{4}27
6. \;\ log_{7}24-log_{7}(y+5)=log_{7}8
7. \;\ 2log_{10}6-\frac{1}{3}log_{10}27=log_{10}x
8. \;\ log_{6}(a^{2}+2)+log_{6}2=2
Hello Ashley. I hope I interpreted correctly.
Try the change of base formula. log_{a}b=\frac{log{b}}{log{a}}
asterisk_man
Apr 12, 2007, 05:05 PM
I'm not sure what is right either galactus. We need ashley to come in and tell us which is right.
for example, to me, the first value should be
\log \left({5^9}\right)
if that's the case she needs to use this formula:
\log_a \left({b^c}\right) = c\log_a \left(b\right)
not saying you're wrong, just thought I'd put the other interpretation in case (s)he wants it.