Logarithms - Intercepts and Inverses

I am going back to college and I'm hoping to test out of a math class. It has been years since HS, so any help would be greatly appreciated.

Determine the intersection of the two logarithmic functions:

y= log.base3(x+25)+3 and
y= log.base3(x+1)+5

I know I need to set them equal to each other to find the point of intersection. But I'm not sure how to solve for x on this one... I know that the +3 and the +5 have to do with the y-int.


Find the inverse function:

y=log.base5(x+6)-2
The -2 is throwing me off. I know I need to switch x and y, then get it back to y=... but I'm not sure how to work it.

The key to these is to get the "log" alone... then switch to exponential form...

y = log (base 3)(x+25) + 3
y -3 = log (base 3) (x+25)
3^(y-3) = x + 25 equation 1
Now the other equation...
y = log (base3)(x+1) + 5
y - 5 = log (base 3)(x+1)
3^(y - 5) = x + 1
Now multiply both sides of this by 9 which is 3^2
(3^2)(3^(y - 5)) = (3^2)( x + 1)
using laws of exponents you get
3^(y-3) = 9x + 9 equation 2
See equation 1 and equation 2 are both equal to 3^(y-3) so they are equal to each other.
x + 25 = 9x + 9
8x = 16
x = 2
Plug that back into an equation anywhere along the line and you will find y = 6
Point of intersection (2,6)


for the inverse..
y + 2 = log(base 5)(x + 6)
change to exponential form
5^(y + 2) = x + 6
Now, switch the x and y to make the inverse...
5^ (x + 2) = y + 6
and solve for y...
y = 5^(x + 2) - 6
voila!!!

log.base3(x+25) + 3 = log.base3(x+1) + 5

log_3(x+25) - log_3(x+1) = 5 - 3

log_3 ( (x+25) / (x+1) ) = 2

3^2 = (x+25) / (x+1)

9 * (x + 1) = x+25

9x + 9 = x + 25

9x - x = 25 - 9

8x = 16

x = 2

y = log_3 ( 2 + 25 ) + 3
y = log_3 ( 27 ) + 3
y ≈ 6

intersection at (2 , 6)

============

y = log_5 (x+6) - 2 ====> switch variables

x = log_5 (y+6) - 2 ===> solve for y

x + 2 = log_5 (y+6)

5^(x + 2) = y + 6

5^(x + 2) - 6 = y = f^-1(x)