Usually when I have a few equation I can figure out how the answer was achieved. I have been tracking my head around this for some time now and I just have only managed to find about 4 different ways to come up with 5.26. My math skills 12 years ago never went past algebra so assume I know nothing more than beginner math when trying to explain this. Lol.
How does 2 to the 2.63 power equal 6.19?
- Thread starter NvrDwn
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2 to the 2.63 power
I know the answer is 6.19 but I have to show work. Mind you I'm nearly 30 never went past algebra in high school.
Usually if I have a full equation I can figure out how it was achieved. But all I've managed to do is come up with 5.26 in 4 different ways.
I know the answer is 6.19 but I have to show work. Mind you I'm nearly 30 never went past algebra in high school.
Usually if I have a full equation I can figure out how it was achieved. But all I've managed to do is come up with 5.26 in 4 different ways.
HallsofIvy
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I'm not sure exactly what your question is. To begin with \(\displaystyle 2^{2.63}\) is not "equal to 6.19". Entering \(\displaystyle 2^{2.63}\) on a calculator, we get 6.1902599741695596201802617784274....That gives 6.19 only if you round to two decimal places. I have no idea how you got "5.26" four different times! In the dark old days "B.C" (Before Calculators) the standard way to calculate \(\displaystyle 2^{2.63}\) would have been to use logarithms: \(\displaystyle log(2^{2.63})= 2.63 log(2)\). Looking up the logarithm of 2 in a "table of logarithms" we find that log(2)= 0.3010 to four decimal places. 2.63 times 0.3010 give 0.7917. Looking up the "anti-logarithm" of 0.7917 in that same table, we find 6.1902.
Another way to do that is to note that \(\displaystyle 2^2= 4\) and \(\displaystyle 2^3= 8\). 2. 63 lies between 2 and 3 so \(\displaystyle 2^{2.63}\) lies between 4 and 8. With a little more work, \(\displaystyle 2^{2.5}= 2^2\sqrt{2}= 4\sqrt{2}= 5.66\) so \(\displaystyle 2^{2.63}\) must lie between 5.66 and 8- but that requires taking the square root which is quite a bit of work unless you use a calculator. With yet more work, \(\displaystyle 2^{2.6}= 2^{2+ 3/5}= 4(2^{3/5})= 4(\sqrt[5]{8})\) but now taking a fifth root is really a chore! Use a calculator!
Another way to do that is to note that \(\displaystyle 2^2= 4\) and \(\displaystyle 2^3= 8\). 2. 63 lies between 2 and 3 so \(\displaystyle 2^{2.63}\) lies between 4 and 8. With a little more work, \(\displaystyle 2^{2.5}= 2^2\sqrt{2}= 4\sqrt{2}= 5.66\) so \(\displaystyle 2^{2.63}\) must lie between 5.66 and 8- but that requires taking the square root which is quite a bit of work unless you use a calculator. With yet more work, \(\displaystyle 2^{2.6}= 2^{2+ 3/5}= 4(2^{3/5})= 4(\sqrt[5]{8})\) but now taking a fifth root is really a chore! Use a calculator!
Harry_the_cat
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2 multiplied by 2.63 = 2 x 2.63 = 5.26
2 to the power of 2.63 is not the same thing. You need to find 22.63 . I am quite sure you would NOT have to do that manually (ie without a calculator).
Do you have a calculator with an xy key? If you press 2 then xy then 2.63 then =, you will get 6.19 (rounded to 2 decimal places).
Note that 22 = 2 x 2 = 4 and 23 = 2 x 2 x 2 = 8. Since 2.63 lies between 2 and 3, then 22.63 will lie between 22 and 23, ie between 4 and 8.
The other answers are correct to tell you that this is best done with a good calculator or logarithm tables. But I suspect that you need to review the laws of exponents, with particular attention to the meaning of fractional exponents. Find some exercises on that topic, and we can help you with those. Purplemath has lessons on exponents from the basic idea to fractional exponents. Using the laws of exponents, we get:
\(\displaystyle 2^{2.63} = 2^2 * 2^{0.63} = 4 * 2^{63/100} = 4 \sqrt[100]{2^{63}}.\)
Now, as Halls said,
\(\displaystyle 2^{0.5} < 2^{0.63} < 2^1 \implies 1.4 < 2^{0.63} < 2 \implies\)
\(\displaystyle 4 * 1.4 < 4 \sqrt[100]{2^{63}} < 4 * 2 \implies 5.6 < 4 \sqrt[100]{2^{63}} < 8.\)
And obviously 5.6 > 5.26.
Between 5.6 and 8 is a very imprecise estimate, but to get a better one by hand involves a fair amount of arithmetic work. A plausible next step would be to estimate 4 times the fourth root of 8.