Solving Logarithmic Equations Involving x & y

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    Logarithmic
In summary, Item #b:Equivalent to \[\log (y) + \log (1000) = \log (y + 1) + \log (x)\]y= (x-3)2 / x, x=e^{ln(x)}
  • #1
Twisted--
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Hi :)
This is my first time using this forum, so thanks for your help in advance.

Homework Statement



Use the laws of logs to write y as a function of x for each. Then state domain.

a) log(xy) = 2 log(x-3)
b) log(y) + 3 = log(y + 1) + log(x)
c) log (x2/y) = 2 log(x + 5)

Answers:
a) y = (x-3)2 / x
b) y = x / (1000 -x)
c) y = x2 / (x + 5)2


The Attempt at a Solution



I got the first one by:

log(xy) = 2 log(x-3)
logx + log y = 2 log(x-3)
log y = 2 log(x-3) - logx
log y = log [ (x-3)2 / x]

I'm not sure about is how they got rid of the "log" to get only x and y in the answer, though. Are you even allowed to divide "log" and cross it out?
I don't know where they got 1000 from in b).
I started by moving all the logs that contain "y" to the left side, but I didn't know where to go from there.
And I'm not sure how to do c).
 
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  • #2
[strike]Recheck the original equation for item b in case you miscopied or misread.[/strike]

Item #c:

On right hand side, [tex] \[
2\log (x + 5) = \log (x + 5)^2
\]
[/tex]

Continuing this from original equation,
[tex] \[
\log \left( {\frac{{x^2 }}{y}} \right) = \log (x + 5)^2
\]
[/tex]

therefore, [tex] \[
\begin{array}{l}
\frac{{x^2 }}{y} = (x + 5)^2 \\
x^2 = y(x + 5)^2 \\
y = \frac{{x^2 }}{{(x + 5)^2 }} \\
\end{array}
\]
[/tex]
 
Last edited:
  • #3
Twisted-- said:
log y = log [ (x-3)2 / x]

I'm not sure about is how they got rid of the "log" to get only x and y in the answer, though. Are you even allowed to divide "log" and cross it out?

No you can't do that. Log is the inverse of exponentiating, in other words, if [tex]y=e^x[/tex] then [tex]x=log_e(y)=ln(y)[/tex]. This means that when you take the exponential of a logarithm, you end up back where you started.
[tex]x=e^{ln(x)}[/tex]

So if you take the exponential of both sides (not the same as dividing both sides by log, which doesn't make any sense) then the logs will cancel.

edit: since the question says to also state the domain, then I guess I should have mentioned this now crucial piece of information. It is not entirely true that [tex]x=e^{ln(x)}[/tex] because logs are only defined for x>0, so you need to place the restriction down that x>0. If you have log(x-5) then x-5>0 or x>5 etc.
 
  • #4
Number b,

Equivalent to [tex]\[
\log (y) + \log (1000) = \log (y + 1) + \log (x)
\]
[/tex]

[tex]\[
\begin{array}{l}
\log (y) - \log (y + 1) = \log (x) - \log 1000 \\
\log \frac{y}{{y + 1}} = \log \frac{x}{{1000}} \\
\frac{y}{{y + 1}} = \frac{x}{{1000}} \\
\end{array}
\]
[/tex]

Continuing with appropriate steps, you obtain the answer shown in your first post.
 
  • #5
If you have f(x)= f(y) with "f" a function you cannot "divide by f"- you divide by numbers not functions. But if you know that a function is "one to one" (and logarithms are) then from "f(x)= f(y)" you can conclude "x= y".
 

What is a logarithmic equation?

A logarithmic equation is an equation in which one of the variables is in the form of a logarithm. Logarithms are the inverse functions of exponential functions, and are commonly used to solve for the unknown variable in exponential equations.

How do I solve a logarithmic equation involving x and y?

To solve a logarithmic equation involving x and y, you must first isolate the logarithm on one side of the equation. Then, you can use the logarithm property logb(xy) = ylogb(x) to rewrite the equation in terms of the base and exponent. Finally, you can solve for the unknown variable by taking the inverse logarithm of both sides.

What are some common logarithmic equations involving x and y?

Some common logarithmic equations involving x and y include logb(x) = y, logb(x + y) = logb(x) + logb(y), and logb(xy) = logb(x) * y. These equations can be solved using the same steps as mentioned above.

What is the importance of solving logarithmic equations involving x and y?

Solving logarithmic equations involving x and y is important in many fields, including mathematics, science, and engineering. These equations allow us to solve for unknown quantities in exponential relationships, and are used in many real-world applications such as calculating population growth, radioactive decay, and sound levels.

Are there any tips for solving logarithmic equations involving x and y?

Some tips for solving logarithmic equations involving x and y include carefully applying logarithm properties, using a calculator or logarithm table for computing values, and checking your solution by plugging it back into the original equation. It is also important to be familiar with the rules of logarithms and practice solving various types of logarithmic equations.

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