Mathematics
18 Online
OpenStudy (anonymous):
2^x=10?
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OpenStudy (anonymous):
the answer is 2^333 repeating b/c it gives u 10.056107. and u basically get 10.
OpenStudy (anonymous):
sorry i didnt explain it so well im not so good at that.
OpenStudy (anonymous):
haha. its fine. I just need to know how to do it with log form
OpenStudy (anonymous):
its for my test tomorrow and no calculator
OpenStudy (nincompoop):
?
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OpenStudy (anonymous):
i need to put 2^x=10 in log form, so log2^10=x
OpenStudy (anonymous):
no calculator?!!!!!! oh wow im sorry. i take trig and we get graphing calcu;ators
OpenStudy (anonymous):
but how would I solve from there?
Parth (parthkohli):
Another way is just to convert from exponential to logarithmic.\[a^b = c \Rightarrow \log_a c = b\]
OpenStudy (nincompoop):
you still have to solve for x
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OpenStudy (anonymous):
thats what i did. but I need to find x
OpenStudy (nincompoop):
log rules :)
Parth (parthkohli):
Just did.
OpenStudy (anonymous):
and I don't know how to do that without a calculator
OpenStudy (nincompoop):
10 = 5+2
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OpenStudy (nincompoop):
does that ring a bell?
OpenStudy (anonymous):
l.o.l.
Parth (parthkohli):
@nincompoop 10 = 5 + 2 = 7? Hao? Pls explan. OP pls.
OpenStudy (nincompoop):
log (10) = log(2) + log (5)
Parth (parthkohli):
k
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OpenStudy (nincompoop):
ni hao ma?
Parth (parthkohli):
I'm good, thanks. @ktmart1n3 got it?
OpenStudy (nincompoop):
log rules :P
OpenStudy (anonymous):
lol. sorta
Parth (parthkohli):
Yeah, log rules!
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OpenStudy (anonymous):
so, my test to solve the equation. and it says 2^x=10 x=? what goes in the question mark space?
OpenStudy (anonymous):
lol. is it going to be log(2)+log(5)?
OpenStudy (nincompoop):
no
OpenStudy (nincompoop):
\[\log_ax=\frac{ \log_bx }{ \log_ba }\]
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Parth (parthkohli):
The above gives a pretty useful result, that is,\[\dfrac{1}{\log_x a}=\log_a x\]
OpenStudy (anonymous):
so, if i want to answer this question for what x is, im going to need my calculator?
Parth (parthkohli):
Do you need the answer in log form?
OpenStudy (nincompoop):
most likely in log form, an approximation would require calculator
OpenStudy (anonymous):
ok
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OpenStudy (anonymous):
What is this shenanigans
Parth (parthkohli):
If so, just use the following:\[a^b = c \Rightarrow \log_a c = b\]
OpenStudy (anonymous):
alright, thank u guys.
Parth (parthkohli):
Just make 'em right substitutions, and you're done with it.
OpenStudy (nincompoop):
again 10 = 5+2 :P