How Do You Simplify E Lnx

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##e^(-ln(x))” ” =” ” 1/x##

##color(brown)(“Total rewrite as changed my mind about pressentation.”)##

##color(blue)(“Preamble:”)##

Consider the generic case of ##” “log_10(a)=b##

Another way of writing this is ##10^b=a##

Suppose ##a=10 ->log_10(10)=b##

##=>10^b=10 => b=1##

So ##color(red)(log_a(a)=1 larr” important example”)##

We are going to use this principle.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Write ##” “e^(-ln(x))” “## as ##” “1/(e^(ln(x))##

Let ##y=e^(ln(x)) =>” “1/y=1/(e^(ln(x))## ………………Equation(1)

……………………………………………………………………………Consider just the denominators and take logs of both sides

##y=e^(ln(x))” ” ->” “ln(y)=ln(e^(ln(x)))##

But for generic case ##ln(s^t) -> tln(s)##

##color(green)(=>ln(y)=ln(x)ln(e))##

But ##log_e(e)” “->” “ln(e)=1 color(red)(larr” from important example”)##

##color(green)(=>ln(y)=ln(x)xx1)##

Thus ##y=x##………………………………………………………………………….

So Equation(1) becomes

##1/y” “=” “1/(e^(ln(x)))” “=” “1/x##

Thus ##e^(-ln(x)) = 1/x##

~~~~~~~~~~~~~~~~~~~~~~~~~~~~##color(blue)(“Footnote”)##

In conclusion the general rule applies: ##” “a^(log_a(x))=x##

 
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How Do You Simplify Lne 4

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##4##

You may put any power in the argument before the ##ln##:##lne^4=4lne## (this goes for ##log##’s to any base)

Then since ##e## is the base of the ##ln##-function (##lnx=log_e x##), it follows that: ##lne=log_e e=1##, just as ##log_10 10=1##

So: ##lne^4=4*lne=4*1=4##

 
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How Do You Solve 2log 2x 1 Loga

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##x=sqrt(10a)/2## (Assuming ##log = log_10##)

##2log_10(2x) = 1+log_10 a##

##2log_10 2x – log_10 a =1##

##2log_10 2x – 2log_10 a^(1/2) =1##

##2log_10((2x)/sqrt(a)) =1##

##log_10((2x)/sqrt(a)) =1/2##

##(2x)/sqrt(a) = 10^(1/2)##

##2x=sqrt(10a)##

##x=sqrt(10a)/2##

 
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How Do You Solve Cost Volume Problems I Know The Equations For Total Revenue And

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how do you solve cost volume problems i know the equations for total revenue and total cost and breakeven etc. but whenever i get an example question of these problems i cant figure them out is there a way you can give me an example of a cost volume problem and I’ll try to solve it and could you walk me through the steps afterwards

 
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How Do You Solve For Equilibrium Price And Quantity For Three Goods Which Are P

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How do you solve for equilibrium price and quantity for three goods which are P^1,P^2,P^3 for demand and supply I’m economics.

Qd1 = 23−5P1 +P2 +P3

and

Qs1 =−8+6P1.

Then :

Qd2 = 15+P1 −3P2 +2P3Qs2 =−11+3P2.

Qd3 = 19+P1 +2P2 −4P3Qs3 =−5+3P3.

 
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How Do You Solve For X In 1 3lnx Ln2 Ln3 3

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The answer is: ##x=27/8e^9##.

First of all we have to add a condition otherwise our equation loses meaning: ##x>0##.

Than:

##1/3lnx+ln2-ln3=3rArrlnx=3(ln3-ln2+3)rArr##

##lnx=3(ln3-ln2+lne^3)rArrlnx=3ln(3e^3/2)rArr##

##lnx=ln(27/8e^9)rArrx=27/8e^9##

(That is positive, so it is acceptable).

 
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How Do You Solve Ln 3x 1 Ln 5 X Ln2

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##x=9##

##ln(3x+1)-ln(5+x)=ln(2)=>## using laws of logs:

##ln[(3x+1)/(x+5)]=ln(2)=>## if ##ln(A)=ln(B)hArrA=B##:

##(3x+1)/(x+5)=2=>## multiply by ##(x+5)##:

##3x+1=2(x+5)=>## expand right side:

##3x+1=2x+10=>## subtract##-2x## and 1 from both sides:

##3x-2x=10-1=>## simplify:

##x=9##

 
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How Do You Solve Ln E X Ln E 3 Ln E 9

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If you use the properties of logarithms, you will need to refer to these:

  1. ##lna^b = blna##
  2. ##lne = 1##

OR

##lne^z = z##

where ##a## and ##b## are constants, ##e = 2.718281828cdots##, and ##z## is either a constant or a variable. Therefore, we can do this:

##lne^x – lne^3 = lne^9##

##lne^x = lne^9 + lne^3##

##xcancel(lne) = 9cancel(lne) + 3cancel(lne)##

##color(blue)(x) = 9 + 3 = color(blue)(12)##

 
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How Do You Solve Ln X 2 16

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I found: ##x=e^8=2,980.95##

We can use the property of the logs that allows you to take out the exponent of the argument and place it as multiplier in front of the log to get:

##2ln(x)=16##

rearrange:

##ln(x)=16/2=8##

use the definition of log to get:

##x=e^8=2,980.95##

 
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How Do You Solve Log 2 Log 4x 1 3

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##x = 5/8##

You used Logarithm of a product (based on the symbol you used which is addition).

##log2 + log(4x-1) = 3####log 2(4x-1) = 3####log (8x-2) = 3####8x-2 = 3####8x = 3+2####8x = 5##

(then divide them with 8 so that we can remove the value of x which is not applicable), so..##8x-:8 = 5-:8##

##x = 5/8##

 
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