someone please help me with these math problems!!..?

Question

Problem 1: Write the expression log a(y+5)+2log a (x+1) as one logarithm. & Problem 2: Solve the equation In(x+5)-In(3)=In(x-3). & Problem 3: Find the exact solution to the equation 3^x+5=9^x.

Answer 1

Do you know your log identities? That would help you here a lot!

Use these log identities to solve your problems:

. log(a) + log(b) = log(ab)
. log(a) - log(b) = log(a/b)
. c*log(a) = log(a^c)

So, for problem 1 (I'm assuming the "a" means that "a" is your base) we get the following:

log a(y+5)+2log a (x+1) =
loga(y+5) + loga((x+1)²) =
loga((y+5)(x+1)²)

Problem 2:

In(x+5)-In(3)=In(x-3)
ln((x+5)/3) = ln(x-3)
(x+5) / 3 = x-3
x + 5 = 3x - 9
2x = 14
x = 7

Problem 3:

3^x+5=9^x

Is 3 raised to the power of x only, or to the power of x+5? It's hard to tell, and it does matter to the solution.

~ ♥ ~

Answer 2

1. log a (y+5)+2log a (x+1)
= log a (y + 5) + log a (x+1)² (as nlog x = log xⁿ)
= log a (x+1)²(y + 5) (as log x + log y = log xy)

2. In(x+5) - In(3) = In(x-3)
So ln((x + 5)/3) = In(x-3)
Taking antilogs
(x + 5)/3 = x - 3
x + 5 = 3x - 9
2x = 14
x = 7

3. 3^x+5=9^x
So 3^x + 5 = 3^(2x)
ie 3^(2x) - 3^x - 5 = 0
Let 3^x = u Thus 3^(2x) = u²
So u² - u - 5 = 0
u = (1 ± √(1 + 20)/2
= (1 + √(21))/2 and the negative solution will be ignored as 3^x > 0 for all real values of x

So 3^x = (1 + √(21))/2
Take logs
log (3^x) = x log3 = log ((1 + √(21))/2)
x = log ((1 + √(21))/2)/log3

If you meant 3^(x + 5) = 9^x
Then 3^(x + 5) = 3^(2x)
Equating indices
x + 5 = 2x
x = 5

Answer 3

for your first problem,
log a(y+5)+2log a (x+1) = log ((y+5)(x+1)^2) ..............to base a

problem 2,
In(x+5)-In(3)=In(x-3)
ln((x+5)/3)=ln (x-3)
(x+5)/3=x-3
x+5=3x-9
x=7
(hope you get the steps)

problem 3,
3^x+5=9^x
let 3^x=t so the above equation simplifies to
t+5=t^2
t^2-t-5=0
now solve for the value of 't' from the above quadratic and then equate it to 3^x to solve for x

Answer 4

log(a)(y + 5) + 2log(a)(x + 1)
log(a)(y + 5) + log(a)((x + 1)^2)
log(a)((y + 5)(x + 1)^2)

if you want this taken further, then

log(a)((y + 5)(x^2 + 2x + 1))
log(a)(x^2y + 2xy + y + 5x^2 + 10x + 5)
log(a)((y + 5)x^2 + 2(y + 10)x + (y + 5))

i would go with log(a)((y + 5)(x^2 + 2x + 1)) as the answer.

-------------------------------------

ln(x + 5) - ln(3) = ln(x - 3)
ln((x + 5)/3) = ln(x - 3)

(x + 5)/3 = (x - 3)
x + 5 = 3(x - 3)
x + 5 = 3x - 9
-2x = -14
x = 7

---------------------------------

3^(x + 5) = 9^x
3^(x + 5) = (3^2)^x
3^(x + 5) = 3^(2x)
x + 5 = 2x
-x = -5
x = 5

Answer 5

1:
log a(y+5) + 2log a(x+1)
=log a(y+5) + log {[a(x+1)]^2}
=log {a(y+5).[a(x+1)]^2}

2:
ln(x+5)-ln(3)=ln(x-3)
ln[(x+5)/3]=ln(x-3)
(x+5)/3=x-3
x=7

3:
3^x+5=9^x
3^x+5=(3^x)^2
(3^x)^2-3^x-5=0
solving this quadratic we get
3^x={1+/-sqrt(21)}/2
take log both sides and divide to get value of x