Solve for x
1.) 4^(2x) - 4^(x+1) - 21 = 0
2.) 4log*base10*(3x - 2) - log*base10*(x+1) = 1 + log*base10*2
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
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