What is the Correct Notation for Logarithmic Equations?

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In summary, the conversation is about simplifying the equation (2^(x+1)) / (5^x)) = 3. The attempt at a solution involved using logarithms to express the left side as a single log term, but the mistake was made of subtracting the two log terms instead of dividing them. The correct approach would have been to divide them, resulting in the equation log (2x+1) - log (5x) = log 3. The confusion may have been caused by the poor notation of log5 x/12, which should be written as log5 (x/12) to avoid ambiguity.
  • #1
ThatDude
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Homework Statement



(2^(x+1)) / (5^x)) = 3


2.The attempt at a solution

(2^(x+1)) / (5^x)) = 3

((log 2 (x+1)) / (x log 5)) = log 3

(x log 2 + log 2) / (x log 5) = log 3

( log (2^ (x+1))) / (log (5^x)) = log 3

log (base: 5^x) (number: 2 ^ (x+1)) = log 3

(5^x) ^ (log 3) = 2 ^ (x+1)

(x log 5) ( log 3) = x log 2 + log 2

x ~ 9.09

I know that this is not the right answer but I don't understand where I went wrong.
 
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  • #2
Hi ThatDude! :smile:

(try using the X2 button just above the Reply box :wink:)
ThatDude said:
(2^(x+1)) / (5^x)) = 3

((log 2 (x+1)) / (x log 5)) = log 3

((log 2 (x+1)) minus (x log 5)) = log 3 :smile:

btw, it would have been simpler to start by expressing the LHS as a multiple of (2/5)x :wink:
 
  • #3
tiny-tim said:
Hi ThatDude! :smile:

(try using the X2 button just above the Reply box :wink:)


((log 2 (x+1)) minus (x log 5)) = log 3 :smile:

btw, it would have been simpler to start by expressing the LHS as a multiple of (2/5)x :wink:

Thank you for helping me out.

I thought that you could only subtract two logs when the number inside the log was being divided. In this case, isn't it the division of two logs, albeit with the same base?

Ex:
= log5 x/12

= log5x - log512

But from what I understand from your post, the following is also true:


log5 x/12 = log5x / log5 12
 
  • #4
ThatDude said:
I thought that you could only subtract two logs when the number inside the log was being divided.

it is

you confused yourself by leaving out a step …

2x+1 / 5x = 3

log (2x+1 / 5x) = log 3

log (2x+1) - log (5x) = log 3 :wink:
 
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  • #5
ThatDude said:
Thank you for helping me out.

I thought that you could only subtract two logs when the number inside the log was being divided. In this case, isn't it the division of two logs, albeit with the same base?

Ex:
= log5 x/12

= log5x - log512

But from what I understand from your post, the following is also true:


log5 x/12 = log5x / log5 12

You have some confusing expressions:

What does log5 x/12 mean?

Is it log (5x/12) or log[itex]_{5}(x/12)[/itex]?
 
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  • #6
ThatDude said:
Thank you for helping me out.

I thought that you could only subtract two logs when the number inside the log was being divided. In this case, isn't it the division of two logs, albeit with the same base?

Ex:
= log5 x/12

= log5x - log512

But from what I understand from your post, the following is also true:


log5 x/12 = log5x / log5 12

It is impossible to say whether you are correct, or not, because of the poor notation. If ##\log 5^{x/12}## means ##\log_5 (x/12)## then what you wrote is true. If it means ##\log \left( 5^{x/12}\right)## then what you wrote is false.
 
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  • #7
tiny-tim said:
it is

you confused yourself by leaving out a step …

2x+1 / 5x = 3

log (2x+1 / 5x) = log 3

log (2x+1) - log (5x) = log 3 :wink:

Ah. I see. Thank you for your help.

Yes... Ray Vickson and SteamKing, it is log5(x/12) ; that is, the base is 5 and the argument is x divided by 12. I apologize for the poor notation.
 
  • #8
ThatDude said:
Ah. I see. Thank you for your help.

Yes... Ray Vickson and SteamKing, it is log5(x/12) ; that is, the base is 5 and the argument is x divided by 12. I apologize for the poor notation.

In that case you should write something like log[5] x/12 or log_5 x/12 or log5 x/12; just about everybody would "get it" if you wrote it in any of these three ways (although it would not hurt to also include a brief verbal description, saying what the '5' means). It would be even better to include parentheses, like this: log[5] (x/12) or log_5 (x/12) or log5 (x/12).
 

1. What is an exponential equation problem?

An exponential equation problem is a mathematical problem that involves variables raised to a power, where the power is a constant value known as the exponent. These types of equations are often used to model growth or decay of a quantity over time.

2. How do you solve an exponential equation problem?

To solve an exponential equation problem, you can use logarithms or the properties of exponents. First, isolate the exponential term on one side of the equation. Then, take the logarithm of both sides (using a base that makes the problem easier to solve). Finally, solve for the variable using algebraic techniques.

3. What are some real-life applications of exponential equations?

Exponential equations are used to model many natural phenomena, such as population growth, radioactive decay, and compound interest. They are also commonly used in fields such as biology, economics, and physics to analyze and predict trends and patterns.

4. What is the difference between an exponential equation and a linear equation?

The main difference between an exponential equation and a linear equation is that the variable in an exponential equation is in the exponent, whereas in a linear equation, the variable is typically in the base. This results in a curved graph for an exponential equation, while a linear equation produces a straight line.

5. Can an exponential equation have a negative exponent?

Yes, an exponential equation can have a negative exponent. This means that the base is being raised to a negative power, which is equivalent to taking the reciprocal of the base raised to the positive power. For example, 2-3 is the same as 1/23 or 1/8.

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