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Mathematics 20 Online
OpenStudy (crashonce):

If x does not equal 0 and 1/x, log10x, x^3 are in GP, which of the following are true? 1) log x = x-1 2)log 10 x= x^2

OpenStudy (crashonce):

@mathmath333

OpenStudy (mathmath333):

what do u know about geometric mean

OpenStudy (mathmath333):

if \(a,b~and~c\) are in geometric progreassion then \(\dfrac{b}{a}=\dfrac{c}{b}\)

OpenStudy (crashonce):

that's correct

OpenStudy (mathmath333):

apply the same to your variables

OpenStudy (crashonce):

how does that help with finding x tho

OpenStudy (mathmath333):

we dont want to find x ,we just have to check if the options are correct or not

OpenStudy (crashonce):

so how do I apply them

OpenStudy (mathmath333):

\(\large\tt \begin{align} \color{black}{\dfrac{1}{x},log 10x, x^3\\~\\ a,b,c }\end{align}\) \(\large\tt \begin{align} \color{black}{a=\dfrac{1}{x},b=log 10x,c= x^3\\~\\ }\end{align}\)

OpenStudy (crashonce):

oh right

OpenStudy (crashonce):

log10x^2 = x^2

OpenStudy (crashonce):

therefore log 10x = x

OpenStudy (mathmath333):

so can u simplify it further

OpenStudy (mathmath333):

\(log(ab)=log(a)+log(b)\)

OpenStudy (crashonce):

1+x^2?

OpenStudy (mathmath333):

what is \[\log{10x} \]

OpenStudy (crashonce):

x

OpenStudy (mathmath333):

\[\log(10x)=\log10+\log x\]btw what is the base of the log

OpenStudy (crashonce):

base 10

OpenStudy (mathmath333):

ok so u got the answer ?

OpenStudy (crashonce):

no I don't get it what is it

OpenStudy (mathmath333):

\[we~found ~that~[\log(10x)=x]\]

OpenStudy (crashonce):

yes but which option confirms it

OpenStudy (mathmath333):

\[\log_{10} 10+\log_{10} x=x\]

OpenStudy (mathmath333):

what is \[\log_{10} 10=\]

OpenStudy (mathmath333):

http://www.wolframalpha.com/input/?i=log_a+a%3D

OpenStudy (crashonce):

1 no I get it but im not sure whichanswer it hints to

OpenStudy (mathmath333):

\(1\)+\(logx\)=\(x\)

OpenStudy (mathmath333):

i think it is equivalent to option \(a.\)

OpenStudy (mathmath333):

i mean option \(1.\)

OpenStudy (jhannybean):

Just solve for \(\log(x)\) now, lol.

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