CPhill

4x /(x^2-2x-15)        factor the denominator as  (x - 5)(x +3)

And since anything that makes the denominator 0, is undefined......then this rational function will be undefined when x = 5 or x = -3

See the graph, here: https://www.desmos.com/calculator/81t61hpgz7

(sqrt of 125)^3x = 625^(2x-1)

Tony, we can solve this one by equatiing bases.......notice that.....625 = 5^4

And sqrt of 125   can be wriiten as  [125]^(1/2) =  [ (5)^3]^(1/2)    =  5^(3/2)

So we have

[ 5^(3/2) ] ^(3x)   =  [5^4 ] ^ (2x - 1)     simplify

5^ [ ( 9/2 ) x] =   5^ [ 8x - 4]

Since we have the bases the same, we can just solve the exponential equations....so we have

(9/2)x = 8x - 4        multiply through by 2

9x = 16x - 8        add 8 to both sides and subtract 9x from both sides

8 = 7x               divide both sides by 7

8 / 7  = x

Thanks, Alan.......your approach makes this much easier.......how did you happen to hit on that particular substitution????

antilog [ (1.1129 ]  =  10^1.1129  =  about  12.9688

There are several but the one most used is :    Area  = [ triangle base * triangle height ]  / 2

Could you give us an example of what you're having trouble with???

(z-3)^4+(z-5)^4=-8   expanding, we have

z^4-12 z^3+54 z^2-108 z+81 + z^4-20 z^3+150 z^2-500 z+625  = - 8        simplify

2 z^4-32 z^3+204 z^2-608 z+714   = 0 

This will factor as :

2 (z^2-8 z+17) (z^2-8 z+21) = 0

Using the quadratic formula to solve the first quadratic factor, we have :

z = 4 - i

z = 4 + i

And using the quadratic formula to solve the second quadratic factor, we have

z =  4 - i sqrt(5) 

z =  4 + i sqrt(5)

The fifth term can be expressed as

1 / [ a + 4d] =  1/28

And the 13th term can be expressed as

1 / [ a + 12d]  = 1/178

So.......this implies that

a + 4d   = 28   (1)       and

a + 12d  =  178      (2)

Multiply (1) by -1 and add it to (2)

8d  = 150        divide both sides by 8

d = 75/4

Using (1)   to find a, we have

a + 4(75/4)   = 28

a + 75  = 28         subtract   75 from each side

a = -47

The first term is given by 1/a   =    1/ - 47  =   -1/ 47

log_b k  = 2.5       means that  b^2.5  = k

log_e b = 2         means that e^2  = b

So  this means that   b^2.5  = k  →   (e^2)^2.5  = k   →   e^5  = k

So 

log_e k   = log_e e^5   =    5 log_e e  =  5 * 1    =  5

1/2(x-3)=5/12 + 2/3(2x-5)        multiply through by 12

6(x -3)  = 5 + 8(2x - 5)     simplify

6x - 18   = 5 + 16x - 40

6x - 18 =  -35 + 16x      add 35 to both sides, subtract 6x from both sides

17  = 10x       divide both sides by 10

17 / 10   = x