equation solving with log?

Question

log(Sub)4 (x^2 - 9) - log(sub)4 (x+3) = 3

solve the equation.

how would I do this one?? please help, explain thanks

Answer 1

log(Sub)4 (x^2 - 9) - log(sub)4 (x+3) = 3

Log subtraction indicates division in the original terms.

log(Sub)4 ((x^2 - 9)/(x +3 ) = 3

log(Sub)4 ((x+3)(x-3)/(x+3)) = 3

log(Sub)4 (x - 3) = 3

4^3 = (x - 3)

x - 3 = 64

x = 67
.

Answer 2

Well I believe with logs :

log (sub x) y = z also equals
x^z = y

Also, log (sub x) y - log (sub x) z = log (sub x) y/z

So applying these two ideas:

log( sub 4) (x^2 - 9)/ (x + 3) = 3

Factor x's:

log (sub 4) (x-3) = 3

Apply first idea:

x-3 = 4 ^3

x-3 = 64

x = 67

Hope that is right!

Answer 3

log(sub)4 ((x^2) - 9) - log(sub)4 (x+3) = 3

Make the left side and the right side the exponent of with a base of four.

4^(log(sub)4 ((x^2) - 9) - log(sub)4 (x+3)) = 4^3

This can be rewritten as:

4^((log(sub)4 ((x^2) - 9))/(log(sub)4 (x+3)) = 64

This can be rewritten as:

(4^(log(sub)4 ((x^2) - 9))/(4^(log(sub)4 (x+3))) = 64

The 4^log(sub)4's cancel.

leaving us with ((x^2)-9)/(x+3) = 64

the top is equal to (x+3)*(x-3), and the (x+3)'s cancel, leaving us with:
x-3 = 64

x = 67

plugging this back into our equation, we get (approximately):

6.0646415 - 3.0646415 = 3

Answer 4

fixing equations with logs and lx.? I actually have a TI-eighty 3 Plus and could prefer to recognize (a thank you to sparkling up) those. a million. sparkling up the equation: Logx + Log(2x + a million) = a million log(x(2x+a million))=a million 2x^2+x=10 2x^2+x-10=0 (2x+5)(x-2)=0 2x+5=0 x=-2.5 x-2=0 x=2 solutions are 2, -2.5 For x=-2.5 as written initially, you will have the logs of two adverse numbers, -2.5 & -4 which oh course are not actual, yet via combining them, you wave the log of a favorable variety, 10 2. sparkling up the equation lnx - lnx(x-3) = ln3 ln(x/(x-3))=ln 3 x/(x-3)=3 x=3(x-3) x=3x-9 -2x=-9 x=9/2=4.5 the respond for a million. is two the respond for 2. is 9/2 or 4.5

Answer 5

since the logs have the same base, they cancel themselves out. so your left with (x^2 - 9)- (x+3)=3

so minus the 3 in the x+3 and so (x^2-9) - (x) = 0

add the 9 to the zero

x^2 - x = 9 <------ so solve that and then you'll have your answer.

Answer 6

OK well first use log base formula to convert to
[log(x^2-9)/(log4)]-[log(x+3)/(log4)]=3.
Then multiply both sides by log4 to get
log(x^2-9)-log(x+3)=3log4.
Change 3log4 to log(4^3) or log 64.
Then change log(x^2-9) to log(x+3)+log(x-3).
The equation so far is.
log(x+3)+log(x-3)-log(x+3)=log64.
The log(x+3)'s are cancelled so log(x-3) = log64 is left. Dropping log from both sides yields x-3=64 so x=67.