Question 60837
1)  Log 3 x = 8
==> log x to the base 3 = 8
Coverting this to exponential form, we have,
x = 3^8
x = 6561.


2) 27^(2x+2) = 81^(x-3)

Now 27 = 3^3  and 81 = 3^4

==> 3^3(2x+2) = 3^4(x-3)
==> 3^(6x+6) = 3^(4x-12)
As the bases are the same,
we equate the powers.
==> 6x+6 = 4x-12
==> 6x + 6 - 4x = 4x - 12 - 4x
==> 2x + 6 = - 12
==> 2x + 6 - 6 = -12 - 6
==> 2x = - 18
==> 2x/2 = -18/2
==> x = - 9



3) LOg 3x+2 to the base 2 = 3
In exponential form this becomes,
3x + 2 = 2^3
==> 3x + 2 = 8
==> 3x + 2 - 2 = 8 - 2
==> 3x = 6
==> 3x/3 = 6/3
==> x = 2


4)e^3x = 80
Taking logarithm on both the side we have..
log[e^3x] = Log 80
==> 3x = log 80  [AS e and log cancel each other]
==> 3x = 1.9031
==> 3x/3 = 1.9031/3
==> x = 0.6344


Good LUck!!!