# Thread: Another confusing Log problems help!

1. ## Another confusing Log problems help!

Solve for x
1.) 4^(2x) - 4^(x+1) - 21 = 0

2.) 4log*base10*(3x - 2) - log*base10*(x+1) = 1 + log*base10*2

2. ## first problem only

Originally Posted by ^_^Engineer_Adam^_^
Solve for x
1.) 4^(2x) - 4^(x+1) - 21 = 0
...
Hello,

rearrange your equation to:

(4^x)^2 - 4*4^x - 21 = 0

That's a quadratic equation in 4^x. So substitute t = 4^x and you'll get:

t^2 - 4t - 21 = 0

Solve this equation for t. (I leave this part for you. There is only one possible solution: 7)

Re-substitute: 7 = 4^x.
Now solve this equation for x = (log(10)7) / (log(10)4) approximately 1.4036...

Greetings

EB

3. Originally Posted by ^_^Engineer_Adam^_^
2.) 4log*base10*(3x - 2) - log*base10*(x+1) = 1 + log*base10*2
Let log denote log_10 (log to the base 10). Then the equation is:

4log(3x - 2) - log(x+1) = 1 + log(2)

Now the first term may be simplified using A log(B)=log(B^A), so

log((3x - 2)^4) - log(x+1) = 1 + log(2),

now using the basic -log(A)=log(1/A) so:

log((3x - 2)^4) + log(1/(x+1)) = 1 + log(2),

and now the addition property of logs log(A)+log(B)=log(AB):

log((3x - 2)^4/(x+1)) = 1 + log(2),

Now log(10)=1 so the right hand side becomes:

log((3x - 2)^4/(x+1)) = log(10)+log(2)=log(20).

Hence:

(3x - 2)^4 / (x+1) = 20.,

which is a quartic, which you must solve using whatever techniques you
have been taught for solving such.

RonL