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OpenStudy (australopithecus):
Log_x(1/2) = -3/2
is the same as
x^(-3/2) = 1/2
OpenStudy (anonymous):
x^(-3/2 = 1/2 .. i think you can try these
OpenStudy (australopithecus):
Logrithmic rules
1) log_b(mn) = log_b(m) + logb(n)
2) log_b(m/n) = log_b(m) – logb(n)
3) log_b(m^(n)) = n · log_b(m)
4) log_3(3) = 1
5) log_e(x) = ln(x)
6) log(x) = y is the same as x =10^(y)
7) log_10(x) = log(x)
OpenStudy (anonymous):
1 / x^3/2 = 1/2 right?
OpenStudy (australopithecus):
yes
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OpenStudy (australopithecus):
x^(-3/2) = 1/2
Raise both sides of the euqation by the power of 2
x^(-6/2) = (1/2)^(2)
x^(-3) = (1/2)^(2)
Now take the negative cube root of both sides
x = (1/2)^(-2/3)
OpenStudy (anonymous):
its not 1/2 its 1/8 .. waa my mistake
OpenStudy (anonymous):
8 = x^3/2
8 = sqrt x^3
64 = x^3
then
x=4?
OpenStudy (australopithecus):
2^(2/3) is the answer if it was 1/8
8^(-1)^(-2/3)
=
8^(2/3)
OpenStudy (anonymous):
yes you are right
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OpenStudy (australopithecus):
simple algebra
OpenStudy (anonymous):
@ampohmeow i think it is
OpenStudy (anonymous):
thanks a lot .. even its simple .. im here to learn
OpenStudy (anonymous):
way to go ampohmeow .. you can post questions here if you like ..
OpenStudy (australopithecus):
Yeah we all have to start somewhere I was having a hard time with this math just 8 months ago now I'm getting 90s on calc I midterms