if log(1234) = 3.0913 , log(769874)=5.88642) , find the value of \[\sqrt[8]{0.000000001234}\]
@satellite73 @FoolForMath
Hint: \[\log_b(y) = x \equiv b^x = y\]
@Hero No idea!
is that first one log base ten?
yes all are with base 10
b = 10
then \[\log(\sqrt[8]{.000000001234})=\frac{1}{8}\log(1234\times 10^{-12})\]
ok
check that i counted the decimal places correctly
u r correct
Is FFM still typing?
aka Across
you get \[\frac{1}{8}\left(\log(1234)+\log(10^{-12})\right)\] \[\frac{1}{8}\left(\log(1234)-12\right)\]
\[ x =\sqrt[8]{0.000000001234}= (1234 \times 10^{-12})^{\large \frac18} \] \[ \implies \log x = \frac 18 \times (\log 1234 -12 ) \]
hard for me to count that high, but if @FoolForMath got the same thing, then i guess it must be right
yup"
so wat is up next!
nothing you are done replace \(\log(1234)\) by 3.0913
i miss across
Plzz help
what do you need?
the answer should be 0.0769874
after solving urs i got -1.11358
I got 0.0769874 as well
then plz show it lol
i don't know. \[\frac{1}{8}\left(\log(1234)-12\right)=\frac{1}{8}\left(3.0913-12\right)\] then a calculator
1/8 of a negative number?
@satellite73, we don't need a calculator, \[\log x = \frac 18 (3.0913-12) = -1.11359 =-2 + 0.88642\] \[\implies x= 10^{-2} \times \text{ Antilog}( 0.88642)=0.0769874 \]
You only came up with that after the fact
?
you are telling me i want to compute \[\log x = \frac 18 (3.0913-12) = -1.11359 =-2 + 0.88642\] with pencil and paper?
Yes, that's what we are supposed to do in exam :)
This is a typical entrance problem.
that is why i do not take exams. but lets say for a minute that you actually managed to do that explain how i am going to compute \[ x= 10^{-2} \times \text{ Antilog}( 0.88642)=0.0769874\]
lol
yeah, without a calculator @FFM
thzz
By reading the question once again.
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