geno3141

log(x + 4) + log(3x + 2)  =  log( (x + 4)(3x + 2) )  =  log( 3x² + 14x + 8 )

log(100) - 2log(2)  =  log(100) - log(2²)  =  log(100) - log(4)  =  log(100/4)  =  log(25)

Combining:  log( 3x² + 14x + 8 )  =  log( 25 )

--->                        3x² + 14x + 8  =  25

--->                        3x² + 14x -17  =  0

--->                      (3x + 17)(x - 1)  =  0

--->     either  3x + 17  =  0            or     x - 1  =  0

                                x  =  -17/3     or           x  =  1

The answer  -17/3  doesn't check; the answer  1  does.

Since  f(-4)  =  -60/13 

and     f(-4)  =  (a/b)x,

then a possibility is:   f(-4)  = (15/13)(-4)   --->   a = 15  and  b = 13

[other possibilities are multiples of  (15/13)]

Let's see what happens with f(4)   --->   f(4)  =  a·b·(4)²  =  (15)·(13)·4²  =  3120.

Since this checks out,  a = 15  and  b = 13.

Step 1:  Mark point "X" on AC at the end of the blue line segment.

Step 2:  BX = DC   --->   BX = ...

Step 3:  XC = BD   --->   XC = ...

Step 4:  AX = AC - XC   --->   AX = ...

Step 5: Use the Pythagorean Theorem on triangle(AXB) to find AB ...

Working backwards:

1)  Looking at the inner f(x):   When will f(x) = 4?

      ---  If x <= 0:  -x + 3  =  4   --->   -x  =  1   --->   x  =  -1

      ---  If x > 0:  2x - 5  =  4   --->   2x  =  9   --->   x  =  4.5

2)  Looking at the outer f(x):  When will f(x)  =  -1?

      ---  If x <= 0:  -x + 3  =  -1   --->   -x  =  -4   --->   x  =  4             (Impossible!  x must be <= 0)

      ---  If x > 0:  2x - 5  =  -1   --->   2x  =  4   --->   x  =  2               (First answer!)

     ---  If x <= 0:  -x + 3  =  4.5   --->   -x  =  1.5   --->   x  =  -1.5     (Second answer!)

      ---  If x > 0:  2x - 5  =  4.5   --->   2x  =  9.5   --->   x  =  4.75     (Third answer!)

Check:  f(f(2))  =  f(-1)  =  4

             f(f(-1.5))  =  f(4.5)  =  4

             f(f(4.75))  =  f(4.5)  =  4

Let  A  =  tan^-1(1/3)   --->   tan(A) = 1/3

Let  B  =  tan^-1(1/7)   --->   tan(B) = 1/7

Let  C  =  tan^-1(1/5)   --->   tan(C) = 1/5

Problem:  tan( A - C + B )   --->   tan( A + B - C )   --->   tan( (A + B) - C )

Since:  tan(X - Y)  =  [ tan(X) - tan(Y) ] / [ 1 + tan(X)·tan(Y) ]

With  X = A + B  and  Y = C   --->    tan( (A + B) - C )

                                                 =     [ tan(A + B) - tan(C) ] / [ 1 + tan(A + B)·tan(C) ]

Since:  tan(X + Y)  =  [ tan(X) + tan(Y) ] / [ 1 - tan(X)·tan(Y) ]

With  X = A  and  Y = B   --->   [ tan(A + B) - tan(C) ] / [ 1 + tan(A + B)·tan(C) ]

=   [ ( tan(A) + tan(B) ) / ( 1 - tan(A)·tan(B) )  - tan(C) ]  /  [1 + ( tan(A) + tan(B) ) / ( 1 - tan(A)·tan(B) )·tan(C) ]

=  [ ( 7 + 3 ) / ( 21 - 1 ) -  (1/5) ]  /  [ 1 + ( 7 + 3 ) / ( 21 - 1 )·(1/5) ]

=  ( 3 / 10) / ( 11 / 21 )

=  3/11

Thanks, EP -- I appreciate your comment!

The measure of angle(A) will be one-half the measure of the intercepted arc:

one-half of 168^o  =   .....

You may want to look here:  https://tutorial.math.lamar.edu/classes/calci/defnofdefiniteintegral.aspx

Since the number 6 was created by 2 * 3, you might want to consider 15 as 3*5.