Do you know the short cut in figuring out these problems?

Question

Can you solve these two problems using a short cut? If so what is the shortcut? The answer to both is 800 (I did them the long way). There must be a short cut, if so let me know.

(2^5)*(5^2)

(10^5)/(5^3)

Answer 1

The shortest way to get the first one is a bit of a trick. Since you have powers of two and five, you can combine them nicely to make tens, which makes life a lot simpler.

(2^5)*(5^2) = 2*2*2*2*2*5*5
= 2*2*2*10*10 (2*5*2*5 = 10*10)
= 8*10*10 (2*2*2 = 8)
= 8*100
= 800

The second expression can be simplified by reducing. First you split the power of ten into powers of two and five:

(10^5)/(5^3) = (2^5*5^5)/(5^3)
= (2^5)*(5^2) (because 5^5/5^3 = 5^2 ... 5*5*5*5*5/(5*5*5) = 5*5)
= 800 (from above)

Hope this helps.

Answer 2

(2^5)*(5^2)
break down the higher power = (2^2)(2^2)(2^1)(5^2)
group the same power = (2^1)((2X2X5)^2)
Do the easy multiplications = 2(20^2) = 2 (400) = 800

(10^5)/(5^3)
Spot the common base and factorize = ((2X5)^5)/(5^3) = (2^5)(5^5)/(5^3)
= (2^5)(5^5)(5^-3)
= (2^5)(5^2)
Do the same as above.

Not sure if this is a shortcut or the long way

Answer 3

Sure.

Write it like this:

(2 * 2 * 2 * 2 * 2) * (5 * 5)
=(2 * 5) * (2 * 5) * (2 * 2 * 2)
=10 * 10 * 8
=100 * 8 = 800

(10 * 10 * 10 * 10 * 10) / (5 * 5 * 5)
=(10/5) * ( 10/5) * (10/5) * 10 * 10
= 2 * 2 * 2 * 10 * 10
=8 * 100
=800

Answer 4

If they were the same base, you could add or subtract the exponents. For example:
(2^5)*(2^2) = 2^7
(10^5)/(10^3) = 10^2

However, if the bases are different, I don't know of any shortcut. If you're thinking of similar bases, that's how. Otherwise, I think your long way was just fine.

Answer 5

These particular problems are easy enough to do in your head, however, a general shortcut is to convert to logarithms. Base 10 are the easiest to use in your head, but natural logarithms work , too.

log (2^5) = 5 log 2
log 2 is approximately 0.3 (.301, but .3 is close enough for quick estimates)

log (5^2) = 2 log 5
log 5 = log 10 - 0.3 (subtracting 0.3 divides by 2, subtracting 0.6 divides by 4, subtracting 0.9 divides by 8)
log 10 = 1

So, I've got:

5 * 0.3 + 2 * 0.7 = 1.5 + 1.4 = 2.9
When converting back to a regular number, the decimal part of the logarithm gives me the 8 (see above) and the 2 tells me how many zeroes to add after the 8.

This is one of the reasons they like to measure the power of signals in decibels. 3 dB = 10 * log 2
Knowing just the log of 2, 4, and 8, you can quickly do power calculations.

Answer 6

For an equilateral triangle, all aspects are the comparable length. Draw an equilateral triangle with one vertex pointing up. Now, draw a vertical line from the top vertex, right down to the backside of the triangle. The vertical line ought to bisect the backside (i.e., it would intersect the middle of the backside). Now, seem at 0.5 of the equilateral triangle, shaped with the help of 0.5 of the backside, the vertical line you in basic terms drew, and between the relax aspects. Use Pythagorean theorem to clean up for the top of the vertical line: - The hypotenuse is 14. - the backside (of the superb triangle) is 14 / 2 = 7. 7^2 + h^2 = 14^2 h = sqrt(14^2 - 7^2) This simplifies right down to h = 7*sqrt(3) Now the area of the equilateral triangle is a million/2 * base * top a million/2 * 14 * 7*sqrt(3) = 40 9*sqrt(3) as quickly as you recognize the thank you to try this, that is reachable to in basic terms keep in mind that the top of an equilateral triangle is comparable to a million/2 * base * sqrt(3).

Answer 7

These problems are short to begin with, so there isn't really a short cut. If you were given much more complicated exponents, logarithms would be the fastest way I can think of.

Let x = (2^5)*(5^2)
log(x) = 5log(2) + 2log(5) = 2.90309...
x = 800

and so on.