The value of

Algebra Level 2

$\large \dfrac{2^{2001} + 2^{1999}}{2^{2000} - 2^{1998}} = \, ?$

2 2000000000+1 10/3 10

1 solution

Sravanth C.
Jan 5, 2016

We can solve this by simply taking out a common factor like this: $\dfrac{2^{2001} + 2^{1999}}{2^{2000}-2^{1998}} = \dfrac{2^{1999}(2^2 + 1)}{2^{1999}(2 - \dfrac 12)}\\ = \dfrac{(5 + 1)}{\dfrac 32} = \boxed{\dfrac{10}3}$

If you feel that the fraction in the denominator is uncomfortable, you can take $2^{1998}$ as the common factor, and simplify it as follows: $\dfrac{2^{2001} + 2^{1999}}{2^{2000}-2^{1998}} = \dfrac{2^{1998}(2^3 + 2)}{2^{1998}(2^2 - 1)}\\ = \dfrac{(8 + 2)}{4 - 1} = \boxed{\dfrac{10}3}$

I solved it similarly, however, I think that it would look more elegantly, if you took $2^{1998}$ before brackets. We would avoid that fraction in denominator.

Zyberg NEE - 5 years, 5 months ago

Thanks for the suggestion, I've edited it accordingly! :)

Sravanth C. - 5 years, 5 months ago

The value of

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