Algebra with huge exponents

**stones44** · December 6th 2007, 07:50 AM

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

Solve for X

**wingless** · December 6th 2007, 07:57 AM

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

**janvdl** · December 6th 2007, 07:59 AM

Originally Posted by stones44

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

Solve for X

$4^{1996} = 4^3 (4^{1993})$

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

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

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

Then we have:

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

$45y = 45(2^x)$

$y = 2^x$

$4^{1993} = 2^x$

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

$x = 3986$

**ThePerfectHacker** · December 6th 2007, 08:00 AM

Originally Posted by stones44

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

Solve for X

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

**janvdl** · December 6th 2007, 08:07 AM

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

**colby2152** · December 6th 2007, 08:21 AM

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!

**janvdl** · December 6th 2007, 08:31 AM

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...

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