# i really need help with math problems please?

if you could show work or explain anything so i can understand that would be great thank you so much

1) if log k= c log v + log p, what is the value of k?

2) what is the solution set of the equation

(x / x - 4) - (1 / x +3) = ( 28 / x^2 - x - 12)

3) for all values of x for which the expression is defined,

(2x + x^2) / (x^2 + 5x + 6) is equivalent to...

4)what is the solution set of the equation x= 2 ([radical] 2x -3)

5) what is the value of x in the equation log [small/low]x 4 + log [small/low]x 9 =2

thank you so much

Relevance

1) if log k= c log v + log p, what is the value of k?

r log a = log (a^r) and log a + log b = log (ab)

combine these to get: log(p* v^c)

so log k = log (p * v^c)

since you have two logs with a common base, the arguments must be the same, so you can drop the logs to arrive at:

k = p * v^c

2) what is the solution set of the equation

(x / x - 4) - (1 / x +3) = ( 28 / x^2 - x - 12)

notice that (x - 4)(x+3) = x^2 - x - 12 = lowest common denominator.

Multiply through by this LCD to get:

x(x+3) - (x - 4) = 28 (multiplying by LCD clears the fraction)

x^2 + 3x - x + 4 = 28 (distribute)

x^2 + 2x - 24 = 0 (combine like terms)

(x + 6) (x - 4) = 0

either x + 6 = 0 or x - 4 = 0

so x = -6 or x = 4

However, if x = 4, that makes the denominator zero in the original equation, so discard this solution. The one solution is: x = -6

3) for all values of x for which the expression is defined,

(2x + x^2) / (x^2 + 5x + 6) is equivalent to...

factor top and bottom:

x(2 + x) /[(x + 3) (x + 2)]

reduce by (2 + x)/(x + 2) (since that = 1)

x / (x + 3) for all x not equal to -2

4)what is the solution set of the equation x= 2 ([radical] 2x -3)

square both sides: x^2 = 4(2x - 3)

distribute: x^2 = 8x - 12

x^2 - 8x + 12 = 0

(x - 6)(x - 2) = 0

x - 6 = 0 or x - 2 = 0

x = 6 or x = 2

check:6 = 2sqrt(2*6 - 3) = 2(sqrt(9)) check

check: 2 = 2sqrt(2*2 - 3) = 2sqrt(1) check

both solutions work (always check to make sure that both algebraic solutions work--they won't always!)

5) what is the value of x in the equation log [small/low]x 4 + log [small/low]x 9 =2

use log a + log b = log (ab)

LHS = log (base x) 36 = 2

convert to exponential (log (base x) 36 = 2 means:

x^2 = 36, or x = 6 (discard x = -6, since a base must be positive)

log (base 6) 4 + log (base 6) 9 = 2 because 6^2 = 36...

x = 6

• 1. logk-logp=clogv

log(k/p)=log(v^c)

k/p=v^c

So, k =p*(v^c)

2. x/(x-4)-1/(x+3)=28/(x^2-x-12)

{x(x+3)-1(x-4)}/{(x-4)(x+3)=28/{(x-4)(x+3)}

x^2+3x-x+4=28

x^2+2x-24=0

(x+6)(x-4)=0

So, x=-6 or 4. But x cannot be 4, because the original equation will be undefined then. So x=-6

3. x(x+2)/{(x+3)(x+2) = x/(x+3)

Or, x^2/4=2x-3 (by squaring)

x^2=8x-12

x^2-8x+12=0

(x-6)(x-2)=0

So, x=6 or 2.

5. log[small/low]x(4*9)=2 (summation of logs of numbers under same base is log of multiplication of the numbers under the same base)

4*9=x^2

x^2=36

x=6