Solving a Logarithmic Equation

[Vertciel]

Hello there,

For the following logarithmic equation, my answer for y was different from the provided answer. Could anyone please see if I have erred?

Thank you.

---

1. If \(\displaystyle (\log_2 x - 2)(log_2 y) = 2\), determine y as a function of x.

My Work:

\(\displaystyle \log_2 y = \frac{2}{\log_2 x - 2}\)

\(\displaystyle \frac{\log y}{\log 2} = \frac{2}{\frac{\log x - 2}{\log 2}}\)

\(\displaystyle \frac{\log y}{\log 2} = \frac{\log x - 2}{\log 4}\)

\(\displaystyle \log y = \frac{\log x - 2 x \log 2}{log 4}\)

\(\displaystyle \log y = \frac{log x - 2}{2}\)

\(\displaystyle \log y^2 = \log x - 2\)

\(\displaystyle y^2 = x - 2\)

\(\displaystyle y = \sqrt{x - 2}\) <-- I discarded the negative solution because y cannot be negative.

However, the provided answer was:

\(\displaystyle y = \frac{\sqrt{x}}{2}\)

 

[stapel]

1. If \(\displaystyle (\log_2 x - 2)(log_2 y) = 2\), determine y as a function of x.

What you have posted meaning the following:

. . . . .(log[sub:1dryqw8f]2[/sub:1dryqw8f](x) - 2)(log[sub:1dryqw8f]2[/sub:1dryqw8f](y)) = 2

Was this what you meant, or is the "-2" supposed to be included within the argument of the first logarithm?

Thank you!

Eliz.

 

[Loren]

\(\displaystyle \log_2y=\frac{2}{\log_2x-2}\)

It appears that your first step was to divide both sides by \(\displaystyle \log_22\) or was that \(\displaystyle \log_{10}2\)?
At any rate compare your method with multiplying both sides by \(\displaystyle \frac{1}{\log 2}\). I think you were in error when you placed the log 2 as the denominator of the denominator on the right side.

 

[Vertciel]

Dear Eliz,

The -2 is part of the argument for the first logarithm: \(\displaystyle \log_2 x - 2\).

Hope this helps.

 

[Vertciel]

\(\displaystyle \log_2y=\frac{2}{\log_2x-2}\)

It appears that your first step was to divide both sides by \(\displaystyle \log_22\) or was that \(\displaystyle \log_{10}2\)?
At any rate compare your method with multiplying both sides by \(\displaystyle \frac{1}{\log 2}\). I think you were in error when you placed the log 2 as the denominator of the denominator on the right side.

Hello Loren,

Thanks for your reply. I divided \(\displaystyle \log_2 x-2\) by \(\displaystyle \log 2\) because I was using the change-of-base formula.

 

[o_O]

You should include brackets around the x - 2 as what you've written means a completely different thing. log[sub:8u3gky24]2[/sub:8u3gky24](x - 2) ? log[sub:8u3gky24]2[/sub:8u3gky24]x - 2.

Anyway, from:

\(\displaystyle \frac{logy}{log2} = \frac{2}{\frac{log(x-2)}{log2}}\)

\(\displaystyle \frac{logy}{log2} = \frac{2log2}{log(x-2)} \quad \mbox{Invert the fraction in the denominator}\)

\(\displaystyle \frac{logy}{log2} = \frac{log4}{log(x-2)} \quad \mbox{Not} \quad \frac{log(x-2)}{log4}\)

It seems that you inverted the fraction for some reason.

As for the provided solution, it is not correct. Consider x = 16, y = 2 (since sqrt{16}/2 = 4/2 = 2).

\(\displaystyle log_{2}(16-2) \cdot log_{2}(2)\)

\(\displaystyle = \frac{log14}{log2} \cdot 1\)

\(\displaystyle \approx 3.8074\)

\(\displaystyle \neq \:2\)

 

[soroban]

Hello, Vertciel!

Evidently, you've misread and mistyped the problem . . .

\(\displaystyle \text{1. If } \log_2(x) - 2\log_2(y) \:= \:2 \text{, determine }y\text{ as a function of }x.\)

\(\displaystyle \text{The provided answer is: }\;y \:= \:\frac{\sqrt{x}}{2}\)

\(\displaystyle \text{We have: }\;\log_2(x) - \log_2(y^2) \;=\;2 \quad\Rightarrow\quad \log_2(y^2) \;=\;\log_2(x) - 2\)

\(\displaystyle \text{Then we have: }\;\log_2(y^2) \;=\;\log_2(x) - \log_2(4)\quad\Rightarrow\quad\log_2(y^2) \;=\;\log_2\left(\frac{x}{4}\right)\)

\(\displaystyle \text{Therefore: }\;y^2\;=\;\frac{x}{4}\quad\Rightarrow\quad y \:=\:\frac{\sqrt{x}}{2}\)