# Writing a logarithm in a form not involving logarithms

• Chijioke
In summary: So, if you want to solve for ##x## or ##y## in terms of ##\beta##, you'll need to use a complex number calculator to do the job.

#### Chijioke

Homework Statement
Write this in a form not involving logarithm.
Relevant Equations
$$\log_x(y)=1/\log_y(x)$$
logyx + logxy = 3/2
Solution
\begin{align*}\log_{ y }{ x } + \log_{ x }{ y } &= \frac{ 3 }{ 2 } \\ \log_{ x }{ y } &= \frac{ \log_{ y }{ y } }{ \log_{ y }{ x } } \\ \log_{ y }{ x } + \frac{ 1 }{ \log_{ y }{ x } } &= \frac{ 3 }{ 2 } \\ \left(\log_{ y }{ x } \right)^ { 2 } + 1 &= \frac{ 3 }{ 2 } \left(\log_{ y }{ x } \right) \\ \left(\log_{ y }{ x } \right) ^ { 2 } &= \frac{ 3 }{ 2 } \left(\log_{ y }{ x }\right) - 1 \end{align*}
What do I do next?

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You have a quadratic equation in $\log_y x$. Solve it.

Chijioke said:
Homework Statement:: Write this in a form not involving logarithm.
Relevant Equations:: $$\log_x(y)=1/\log_y(x)$$

logyx + logxy = 3/2
Solution
\begin{align*}\log_{ y }{ x } + \log_{ x }{ y } &= \frac{ 3 }{ 2 } \\ \log_{ x }{ y } &= \frac{ \log_{ y }{ y } }{ \log_{ y }{ x } } \\ \log_{ y }{ x } + \frac{ 1 }{ \log_{ y }{ x } } &= \frac{ 3 }{ 2 } \\ \left(\log_{ y }{ x } \right)^ { 2 } + 1 &= \frac{ 3 }{ 2 } \left(\log_{ y }{ x } \right) \\ \left(\log_{ y }{ x } \right) ^ { 2 } &= \frac{ 3 }{ 2 } \left(\log_{ y }{ x }\right) - 1 \end{align*}
What do I do next?
Hello @Chijioke.

I'm pretty sure that the problem as stated has no solution.

However,

##\displaystyle \ \log_y x - \log_x y =\frac 3 2 \ ##

does have a solution.

scottdave, Chijioke and PeroK
I was not asked to find the value of x and y. I was only asked to write it in a form not involving logarithm. Is writing it in a form not involving logarithm the same as finding the value of x and y or solving the equation persay?

Chijioke said:
I was not asked to find the value of x and y. I was only asked to write it in a form not involving logarithm. Is writing it in a form not involving logarithm the same as finding the value of x and y or solving the equation persay?
"Solving" the quadratic equation, as suggested by @pasmith, will not give specific values for ##x## and/or ##y##. What the solution gives is possible values for ##\displaystyle \log_y x## or equivalently, values for ##\displaystyle \log_x y## .

Let's say you get the result ##\displaystyle \beta= \log_y x##, for some real number, ##\displaystyle \beta##. Writing that in exponential form gives you:

##\displaystyle x=y^{\,\beta}##

A difficulty with the problem, as it is written, is that the solutions for ##\beta## are complex numbers, that is, they have an imaginary part.

Chijioke and PeroK