I have a quick question (terms of powers of 10) if you have 2^90=(2^10)^9=(10^9)^9? is this last part correct or what do I do
I don't believe the last part is consistent with the first two. You changed bases, going from 2 to a 10. The last part (10^9)^9 is a much larger quantity than (2^10)^9
so how do I change this to the power of 10 from the 2nd point?
what more do you want than \(\large 2^{90}=(2^{10})^9\) ?
I don't take it any further?
not as far as I can see
even if I have to change it to a power of 10
you need to change it to base 10 ? i.e. 10^(something) ?
yes
maybe we can use a change of base log formula somehow... are you working with logarithms?
exponential functions
This was not made clear in the question, please state what the problem states.
estimate each quantity in terms of powers of ten. A) 2^90 B)4^50
estimate! that changes everything!
oh sorry
I have \[(2^{10})^{9}\] Then I am not sure what to do
I'm not so good at estimating but from y calculator A. 1.2 10^27 and B. 1.2 10^30
yeah, I know of know method for this either I used a calculator as well
how did you input this into a calc
I rounded it off to the 1.2 and the calculator gave the power of 10
ok I will do that with the rest thanks so much
read the part that says "scientific notation" in the link and it gives the power of ten
Use the\[x ^{y}\]key
you're welcome
I found that thanks
For my calculator I would enter 2 then press the x^y key,then press 27, then equal
i have a TI 84
@Danyel, do you have that function on your calculator?
I don't see it
You could use google or Wolfram
I just went in to there wolfram what a cool site I didn't even know it existed thanks a tons
I just used google, you just put 2^90 in the search window and you will get the results. Try it you may like it.
Yes, Wolfram is great, especially for solving quadratics!
I wish I had this earlier this month I really could have used it.
Good luck with your studies.
Thanks for all your help
You're certainly welcome
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