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Mathematics 16 Online
OpenStudy (diogop110):

How can I solve this without using calculator? x=(10^-2.88)²/0.1

OpenStudy (alivejeremy):

Need help?

OpenStudy (diogop110):

Yes, please.

OpenStudy (alivejeremy):

Did you right the question wrong tho?

OpenStudy (diogop110):

did I write the question wrong? No. It's right...

OpenStudy (alivejeremy):

Just double check

OpenStudy (diogop110):

ok!

OpenStudy (daniel.ohearn1):

If it's 10^-2.88

OpenStudy (diogop110):

I came this far: x=10^-4,76. But I need to get to 1,7x10^-5

OpenStudy (alivejeremy):

Can you screenshot it? :)

OpenStudy (diogop110):

Just a sec

OpenStudy (diogop110):

OpenStudy (freckles):

you can write -4.76 as -5+.24 (in sorry I can't bring myself to use a comma for the decimal point for some reason but I know what you mean by the comma ) so you will need to show 10^(.24) is approximately 1.7

OpenStudy (diogop110):

Oh, @freckles that's alright...my mistake. In my country we use comma for the decimal...I forgot to change

OpenStudy (freckles):

yea i know it is like that in some countries :p it is cool

OpenStudy (diogop110):

@freckles how can I show that?

OpenStudy (daniel.ohearn1):

First thing is first, by PEMDAS, 10^-2.88 is 1/(10^2.88) is 1/(10^2^(1.44)) is one way to look at it.. The other way is 1/(10^2.88) is 1/(10^2*10^.88) either way you need to have good knowledge of the graph of y=10^x or log base 10 of .88 or 1.44

OpenStudy (diogop110):

@daniel.ohearn1 so I can't solve this without a graph or calculator?

OpenStudy (daniel.ohearn1):

You can but it's tricky. log base 10 of 1.44 is ln(1.44) / ln(10) somewhere you need to be like okay, I give a good estimate of this value just using my mind.

OpenStudy (diogop110):

@daniel.ohearn1 well, thanks. It's part of a question about ionic equilibrium, so... I think I'm gonna use calculator. Thank you.

OpenStudy (daniel.ohearn1):

if you can answer what power must e be taken to get 1.44

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