If the log is expressed with base 9 it is equal to:

  • Thread starter Jaco Viljoen
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In summary, the equation log3(x-3)-log1/3(x+1) is equal to log9(x-3) / log93 - log9(x+1) / log91 / 3.
  • #1
Jaco Viljoen
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9

Homework Statement


If log3(x-3)-log1/3(x+1) is expressed in terms of a logarithm with base 9, then it is equal to?

Homework Equations


logba=logca/logcb

The Attempt at a Solution


I know I need to get rid of the 1/2 by multiplying but can't remember how to apply it.

log3(x-3)-log1/3
=log9(x-3) / log93 - log9(x+1) / log91 / 3
= log9(x-3) / 1/2 - log9(x+1) / -1/2

Can someone tell me how to display fractions correctly?
Thank you.

Jaco
 
Last edited:
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  • #2
Jaco Viljoen said:
log9(x+1) / log91 / 3
where'd you get that ##(x + 1)## i don't see it in the question..
Jaco Viljoen said:
log9(x-3) / 1/2 - log9(x+1) / -1/2
Well, you finished the hard part,
i might be wrong , but can't you just takr ##2## common and express the the 2 term expression in terms of ##log_9\frac{numerator}{denominator}##
Jaco Viljoen said:
Can someone tell me how to display fractions correctly?
if you're referring to typing like this..
$$\frac{log_9(x-3)}{\frac{1}{2}} - \frac{log_9(x+1)}{\frac{-1}{2}}$$this might help... go through this https://www.physicsforums.com/threads/physics-forums-faq-and-howto.617567/#post-3977517
 
Last edited:
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  • #3
Suraj M said:
where'd you get that ##(x + 1)## i don't see it in the question..
Its in the question, I struggle to read in the code, sorry about that

Suraj M said:
Well, you finished the hard part,
i might be wrong , but can't you just takr ##2## common and express the the 2 term expression in terms of ##log_9\frac{numerator}{denominator}##

This is what it feels like I should do, but I am not sure? would I add a 2 in from of the log?
Suraj M said:
if you're referring to typing like this..
$$\frac{log_9(x-3)}{\frac{1}{2}} - \frac{log_9(x+1)}{\frac{-1}{2}}$$this might help... go through this https://www.physicsforums.com/threads/physics-forums-faq-and-howto.617567/#post-3977517

Yes that's what I am referring to, Thank you.
 
  • #4
Jaco Viljoen said:

Homework Statement


If log3(x-3)-log1/3(x+1) is expressed in terms of a logarithm with base 9, then it is equal to?

Homework Equations


logba=logca/logcb

The Attempt at a Solution


I know I need to get rid of the 1/2 by multiplying but can't remember how to apply it.

log3(x-3)-log1/3
⇒ log9(x-3) / log93 - log9(x+1) / log91 / 3
⇒ log9(x-3) / 1/2 - log9(x+1) / -1/2
The ⇒ means "implies", which isn't appropriate here. What you should have above is =. The expression on the first line is equal to the one on the second line, which in turn is equal to the one on the third line.

The formula you show as your relevant equation I don't keep in memory. Instead I derive it when I need it, which takes a little longer, but so be it.

Here's an example of how I would rewrite ##log_2(x + 5)## as a logarithm in base 4.
Let ##y = log_2(x + 5)##
Then ##x + 5 = 2^y## See comment 1 below.
Taking the log (base 4) of both sides, I get
##log_4(x + 5) = log_4(2^y)##
Or, ##log_4(x + 5) = y * log_4(2)##, using one of the properties of logs
Solving for y, I get
##y = \frac{log_4(x + 5)}{log_4(2)} = \frac{log_4(x + 5)}{1/2} = 2 * log_4(x + 5)## See comment 2 below
Then ##log_2(x + 5) = 2 * log_4(x + 5)##

1. I used the most important concept in logarithms; that a logarithm is an exponent on whatever the base happens to be. For example, ##log_2(x + 5)## means the exponent on 2 (the base) the results in x + 5.
2. In this step I replaced ##log_4(2)## with 1/2. ##log_4(2)## means the exponent on 4 that results in 2. In other words, ##4^? = 2##. The exponent has to be 1/2, since ##4^{1/2} = 2##.

Jaco Viljoen said:
Can someone tell me how to display fractions correctly?
 
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  • #5
Mark44 said:
Here's an example of how I would rewrite ##log_2(x + 5)## as a logarithm in base 4.
Let ##y = log_2(x + 5)##

Then ##x + 5 = 2^y## See comment 1 below.
Taking the log (base 4) of both sides, I get

##log_4(x + 5) = log_4(2^y)## Why is there log=log now?
Or, ##log_4(x + 5) = y * log_4(2)##, using one of the properties of logs
Solving for y, I get
##y = \frac{log_4(x + 5)}{log_4(2)} = \frac{log_4(x + 5)}{1/2}##
Is this not what I have done on mine?

Mark44 said:
##\frac{log_4(x + 5)}{1/2}= 2 * log_4(x + 5)##
Then ##log_2(x + 5) = 2 * log_4(x + 5)##

Could you explain the step above to me?*2 to remove the 1/2, but why does the 2 remain in front of the log?
Because this is what I want to do but I am not sure why...[/QUOTE]
 
Last edited:
  • #6
Mark44 said:
Here's an example of how I would rewrite ##log_2(x + 5)## as a logarithm in base 4.
Let ##y = log_2(x + 5)##
Jaco Viljoen said:
how is there log=log now?
I don't know what you mean. What are you asking here?
Mark44 said:
Then ##x + 5 = 2^y## See comment 1 below.
Taking the log (base 4) of both sides, I get
##log_4(x + 5) = log_4(2^y)##
Or, ##log_4(x + 5) = y * log_4(2)##, using one of the properties of logs
Solving for y, I get
##y = \frac{log_4(x + 5)}{log_4(2)} = \frac{log_4(x + 5)}{1/2}##

Jaco Viljoen said:
Is this not what I have done on mine?
Yes, this is what you did. You said that you needed to get rid of the 1/2 by multiplying, but couldn't remember how to do it. I wasn't sure if you knew what you were doing, or were just following a cookbook formula that you might not remember or might remember incorrectly. What I wrote was an explanation of what should happen.
Mark44 said:
##\frac{log_4(x + 5)}{1/2}= 2 * log_4(x + 5)##
Then ##log_2(x + 5) = 2 * log_4(x + 5)##

Jaco Viljoen said:
Could you explain the step above to me?*2 to remove the 1/2, but why does the 2 remain in front of the log?
Because this is what I want to do but I am not sure why...
Divide by 1/2 is exactly the same as multiplying by the reciprocal of 1/2 -- in other words, the same as multiplying by 2.
 

Related to If the log is expressed with base 9 it is equal to:

What does it mean to express a log with base 9?

Expressing a log with base 9 means that the logarithmic function is calculated using 9 as the base number. This base number determines the relationship between the input and the output of the logarithmic function.

How do you solve for the value of a logarithm with base 9?

To solve for the value of a logarithm with base 9, you can use the formula log9(x) = y, where x is the input and y is the output. You can also use a calculator or a logarithm table to find the value.

What is the difference between a logarithm with base 9 and a natural logarithm?

The main difference between a logarithm with base 9 and a natural logarithm is the base number used. A natural logarithm uses the irrational number e (approximately 2.718) as the base, while a logarithm with base 9 uses the number 9 as the base.

How is the value of a logarithm with base 9 affected by the input?

The value of a logarithm with base 9 is affected by the input through the inverse relationship between the base and the exponent. As the input increases, the value of the logarithm will also increase, but at a slower rate.

What are some real-life applications of using logs expressed with base 9?

Logs expressed with base 9 can be used in various fields such as finance, chemistry, and physics. For example, in finance, logarithms with base 9 can help calculate compound interest or stock returns. In chemistry, they can be used to measure pH levels, while in physics, they can be used to calculate decibels and earthquake magnitudes.

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