Another Algebra 2 Log Question! But this time Ln! 10 Points!?

Question

Ok, so I again can't solve another problem and am starting to get worried... XD
ln (x – 2) + ln (x + 2) = ln 5

When I solve it I always get either a negative and therefore non-real answer or 3.
When I plug 3 back into the equation: BAM! Doesn't work.

This time I'm gonna show my work so if possible you peeps can write a critique... :D
Here's how I fail:
ln (x – 2) + ln (x + 2) = ln 5
Now I multiply inside the parenthesis...
ln(x^2-4)=ln(5)
Divide both sides by ln()
x^2-4=5
add 4 to both sides
x^2=9
square root of both sides
x=3

Checking my work:
ln (3 – 2) + ln (3 + 2) = ln 5
1.609436912=1.609436912

Wait what? I got it right... but it was wrong just a minute ago...
XD

Ok peeps, instead of answering that for 10 points answer this:
What color is Itachi Uchiha's ring?

Answer 1

Ha! OK, I guess scarlet?





EDIT: You're kidding!? That was a guess! Smiles for me!





Now to the problem:



1/2log x – log x = log 2



Using log properties we can make this



1[(logx)/(2)] = log(2)



log(x) = -2log(2)



log(x) = log(1/4)



x = 1/4



:)

Answer 2

ln is just log to base e, so all the same rules apply.



ln(a) + ln(b) = ln(a * b)



When I get confused, I try to think in powers of 10.



log(100) + log(1000) = 2 + 3 = 5 = log(100000), 100 * 1000 = 100000



ln(x - 2) + ln(x + 2) = ln(5)

ln((x - 2) * (x + 2)) = ln(5)

(x - 2) * (x + 2) = 5

x^2 - 4 = 5

x^2 = 9

x = +/- 3



However, if x = -3, ln(x - 2) = ln(-5), which is not allowed. So x = 3.



***



(1/2)log x – log x = log 2

log(x^(1/2)) - log(x) = log(2)

log((x^(1/2)/x) = log(2)

x^(1/2) / x = 2

x^(1/2) = 2x

x = 4x^2

4x^2 - x = 0

x * (4x - 1) = 0

x = 0 or 4x - 1 = 0

x = 0 or x = 1/4



Since you can't take log of zero, x = 1/4



Remember that squaring a function as we've done can introduce invalid solutions, so let's verify our answer.



(1/2) * log(1/4) - log(1/4) = (1/2) * -0.6021 + 0.6021 = 0.3010 = log(2)



So that's a valid solution.

Answer 3

Just a note you are not dividing by the ln() to clear the logs...



if the ln's are equal, the expressions are equal.



1/2 log x - log x = log 2



log x^1/2 - log x = log 2



log x^1/2/x = log 2



log x^-1/2 = log 2



x^-1/2 = 2

square both sides

x^-1=4

x=1/4

Answer 4

the logo is ln (the letters LN, in lowercase) - it stands for "organic logarithm"; the backside for those logs is a value that we call "e" (this is a precise value that demands an infinited numebr of digits, so this is greater convenient to easily use the letter "e" to represent the cost). in case you have a quantity N, and the quantity is comparable to e^x then, by technique of definition, ln(N) = x The organic log (ln) of a quantity (N), is the exponent which you're able to could use with e, as a manner to represent that quantity. the regulations of logarithms are the comparable because of the fact the regulations for exponents. To multiply values written as exponents of an identical base, upload the exponents. to represent the log of a multiplication, upload the logs of each ingredient. (this is the comparable rule, even even with the incontrovertible fact that the words sound diverse). to advance a capacity to a diverse capacity, multiply the powers. For the log of a capacity to a diverse capacity, multiply the log by technique of the capacity. a million. the unique question could have been: ln(a) - 2*ln(b) + (a million/2)*ln(c) using the 2d rule in opposite, on the final term: ln(a) - 2*ln(b) + ln[c^(a million/2)] using the 2d rule in opposite, on the 2d term ln(a) + ln[b^(-2)] + ln[c^(a million/2)] b^(-2) is the comparable as a million/(b^2) (a detrimental exponent declares a fraction) c^(a million/2) = ?c (if the capacity is a fraction, it declares a root) ln(a) + ln(a million/b^2) + ln(?c) using the 1st rule in opposite (sum of logs = log of product) ln[ a?c / b^2 ] ----- 3 - 5*ln(3) ln(3) is in basic terms a value. this is many times got here upon by technique of moving into 3 interior the calculator, then pressing the biggest marked LN (or ln). On my calculator, i'm getting a million.09861... then you definately multiply that by technique of five to get 5.49306... This value gets subtracted from 3 3 - 5*ln(3) 3 - 5*(a million.09861...) 3 - 5.49306... -2.49306... 3. it rather is an equation the place you're able to detect a value of x For that, you employ the definition on the initiating of my answer for log The quantity N is e^(2x) for this reason, the organic log of N could be 2x e^(2x) = 12 take the organic go browsing the two sides 2x = ln(12) 2x = 2.4849... divide the two sides by technique of two

Answer 5

Try this:

ln (x -2) + ln (x+2) = ln( (x-2)(x+2) ) = ln (x^2 - 4) = ln (5)



take the e of both sides, you get:

x^2 - 4 = 5

x^2 = 9

x = +/- 3



Can only use x = +3



Check the work:

ln (3 - 2) + ln (3 + 2) = ln(1) + ln(5) = ln(5), since ln(1) = 0

Answer 6

Red.

Answer 7

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