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A few basic C2 log questions i cant seem to do.. help would be much appreciated

9) Given that y = 10^x, show that:
a) y^2 = 100
easy.. (10^x)^2 = (10^2)x = 100^x
b) y/10 = 10^x-1
any ideas how to show this one?

14) Given that p = log_q 16 (q as the base), express the following in terms of p.
a) log_q 2 don't know how to do this
b) log_q 8q don't know how to do this either :frown:

3) Solve, giving your answers as exact fractions, these simultaneous equations:

8^y = 4^(2x+3)
log_2 y = 4 + log_2 x
equation 1: (2^3)y = (2^2)^(2x+3)
2^3y = 2^(4x+6)
3y = 4x + 6
I've probably already gone wrong here. If I have, what was i supposed to do? and If i haven't, where do I go from here?

THANK YOU! ANY help is appreciated. this is a chapter im struggling on quite a bit :frown:
Reply 1
Avatar for maxfire
maxfire
14
dumb maths student
Question.

9b.
y=10xy=10^x
y10=10x10=10−110x=10x−1\frac{y}{10}=\frac{10^x}{10}=10^{-1}10^x = 10^{x-1}

14a
Rules of logarithms:
logqab=blogqalog_qa^b = blog_qa which you can apply to get blog2, and put it in terms of p.

14b
log(ab)=log(a)+log(b)log(ab) = log(a) + log(b)
apply it to log(8q)
log8q=log(8)+log(q)log8q = log(8) + log(q)
logqq=1log_qq=1 and you can use the rule in 14a as well to get an answer.

3
22y=24x+62^{2y}=2^{4x+6}
therefore:
2y=4x+62y=4x+6
y=2x+3y=2x+3
Equation 2:
log2(2x+3)=4+log2xlog_2(2x+3)=4+log_2x
log2(2x+3)=log216+log2xlog_2(2x+3)=log_216+log_2x
log2(2x+3)=log216xlog_2(2x+3)=log_216x
2x+3=16x2x+3=16x

Then solve.
Ask if you need more help :smile:

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