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Mathematics 15 Online
OpenStudy (anonymous):

If Logx (1 / 8) = - 3 / 2, then x is equal to

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

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

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

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