Writing a logarithm in a form not involving logarithms

[Chijioke]

Homework Statement

Write this in a form not involving logarithm.

Relevant Equations

$$\log_x(y)=1/\log_y(x)$$

log_yx + log_xy = 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?

 

[pasmith]

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

 

[SammyS]

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

log_yx + log_xy = 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.

 

[Chijioke]

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?

 

[SammyS]

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.