How is:
$ \log_{(xyz)^{1/3}} \left(\frac{yz}{x^{3}}\right)^{2}$
expressed in terms of a and b when:
$ a = \log_{y} x $
$ b = \log_{z} x $
EDIT: It is z in the numerator, apologies i posted by memory. Also reframed the question.
Asked
Active
Viewed 65 times
1
-
1Are you sure that it is $\frac{yx}{x^3}$ in the $\log$ – Nikunj Feb 24 '16 at 17:55
-
1Rewrite all expressions using the natural logarithm only. – Feb 24 '16 at 17:56
-
1there is no equation – Dr. Sonnhard Graubner Feb 24 '16 at 18:01
-
This is all the information I was given. – McTaffy Feb 24 '16 at 18:01
-
Hint: Use the property $a = \frac{logx}{logy}$ and $b = \frac{logx}{logz}$ – Apoorv Feb 24 '16 at 18:03
-
Thanks, I didn't now the two laws at the beginning about the index of a base, and divide by the log of the base. – McTaffy Feb 24 '16 at 18:15
2 Answers
1
$$\log _{ (xyz)^{ \frac { 1 }{ 3 } } } \left( \frac { yz }{ x^{ 3 } } \right) ^{ 2 }=\frac { 2 }{ \frac { 1 }{ 3 } } \log _{ (xyz) } \left( \frac { yz }{ x^{ 3 } } \right) =6\frac { \log _{ x }{ \left( \frac { yz }{ x^{ 3 } } \right) } }{ \log _{ x }{ \left( xyz \right) } } =6\frac { \log _{ x }{ \left( y \right) +\log _{ x }{ \left( z \right) } -3\log _{ x }{ x } } }{ \log _{ x }{ \left( x \right) +\log _{ x }{ \left( y \right) } +\log _{ x }{ \left( z \right) } } } =\\ =6\frac { \log _{ x }{ \left( y \right) +\log _{ x }{ \left( z \right) } -3 } }{ 1+\log _{ x }{ \left( y \right) } +\log _{ x }{ \left( z \right) } } =6\frac { \frac { 1 }{ \log _{ y }{ \left( x \right) } } +\frac { 1 }{ \log _{ z }{ \left( x \right) } } -3 }{ 1+\frac { 1 }{ \log _{ y }{ \left( x \right) } } +\frac { 1 }{ \log _{ z }{ \left( x \right) } } } =6\frac { \frac { 1 }{ a } +\frac { 1 }{ b } -3 }{ 1+\frac { 1 }{ a } +\frac { 1 }{ b } } =6\frac { \frac { a+b-3ba }{ ba } }{ \frac { ab+a+b }{ ab } } =\\ =\frac { 6a+6b-18ba }{ ab+a+b } $$
haqnatural
- 21,578
-
I think it is not really helpful for the OP to post the entire solution. – Friedrich Philipp Feb 24 '16 at 18:07
1
This can be written as, $$\frac{\log (\frac{yz}{x^3})^2}{\log (xyz)^{1/3}}$$ =$$\frac{6\log y+6\log z-18\log x}{\log x+ \log y + \log z}$$ Now given that $$a=\frac {\log x}{\log y},b=\frac{\log x}{log z}$$ we get $\log x=a \log y $ and $\log z= \frac{a}{b}\log y$ Now simply substitute the values of $\log x$ and $\log z$ in the expression to get $$\frac{6\log y+6\frac{a}{b}\log y- 18a \log y}{a\log y+\log y+\frac{a}{b}\log y}$$ =$$\frac{6b+6a-18ab}{ab+b+a}$$
Nikunj
- 6,160