# Thread: Algebra with huge exponents

1. ## Algebra with huge exponents

4^1996 - 4^1995 - 4^1994 + 4^1993 = 45(2^X)

Solve for X

2. 4^1996 - 4^1995 - 4^1994 + 4^1993 = 45(2^x)
4^1993(4^3 - 4^2 - 4^1 + 4^0) = 45(2^x)
2^3986*45 = 45*(2^x)
x = 3986

3. Originally Posted by stones44
4^1996 - 4^1995 - 4^1994 + 4^1993 = 45(2^X)

Solve for X
$\displaystyle 4^{1996} = 4^3 (4^{1993})$

$\displaystyle 4^{1995} = 4^2 (4^{1993})$

$\displaystyle 4^{1994} = 4 (4^{1993})$

Set $\displaystyle y = (4^{1993})$

Then we have:

$\displaystyle 64y - 16y - 4y + y = 45(2^x)$

$\displaystyle 45y = 45(2^x)$

$\displaystyle y = 2^x$

$\displaystyle 4^{1993} = 2^x$

$\displaystyle 2^{2 \times 1993} = 2^x$

$\displaystyle x = 3986$

4. Originally Posted by stones44
4^1996 - 4^1995 - 4^1994 + 4^1993 = 45(2^X)

Solve for X
$\displaystyle 4^{1996}-4^{1995}-4^{1994}+4^{1993} = 4^{1993}(4^3 - 4^2 - 4 + 1) = 45\cdot 2^x$
Thus,
$\displaystyle 45 \cdot 4^{1993} = 45 \cdot 2^x \implies 4^{1993} = 2^x$.
Now convert both sides to base 2 and finish the problem.

5. Hehehe, 3 answers to one problem. At least all of them are right.

6. Originally Posted by janvdl
Hehehe, 3 answers to one problem. At least all of them are right.
That was great guys -- working together in harmony!

7. Originally Posted by colby2152
That was great guys -- working together in harmony!
I would say it was simply all of us posting at the same time...