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Mathematics 15 Online
OpenStudy (anonymous):

Hi everyone.I need help about logarithmic functions.. find x without using a table log sub10 x

OpenStudy (amistre64):

log(x) = y

OpenStudy (amistre64):

what is it you are needing with the helping?

OpenStudy (anonymous):

how to find x without using a table

OpenStudy (amistre64):

x is an independant variable; there is nothing to find;

OpenStudy (amistre64):

x = 3, x = 12, x = 16543463, x = .34, x = nothing 0 and below

OpenStudy (anonymous):

ah sorry the given is lacking

OpenStudy (anonymous):

log sub10 x=2

OpenStudy (amistre64):

ahh lol in other words, logs point to exponents; log10(x) = 2 means: 10^2 = x

OpenStudy (anonymous):

\[\log_{10} x=1 .....................means................x=10^{1}\]

OpenStudy (amistre64):

or do you mean; what does log10(2) = ? then 10^? = 2

OpenStudy (anonymous):

\[\log_{10} x=2\]

OpenStudy (amistre64):

and for the record: "log" by itself means; log(base 10)

OpenStudy (amistre64):

log(x) = 2

OpenStudy (amistre64):

then 10^2 = x; x = 10*10 x = 100

OpenStudy (anonymous):

amistre64 what if the base is not 10....

OpenStudy (amistre64):

if it aint base 10 then we sullpy the base. say base 7 log7(x) = 2 7^2 = x 49 = x

OpenStudy (amistre64):

logs point to exponents; they are inverses of each other

OpenStudy (anonymous):

what if the given is like this lnx=ln2+ln3+ln

OpenStudy (amistre64):

\[log_7(x)=2\] \[7^{log_7(x)}=7^2\] \[x = 7^2\] \[x = 49\]

OpenStudy (amistre64):

lna + lnb = ln(ab)

OpenStudy (anonymous):

can you explain further

OpenStudy (amistre64):

\[ln(x)=ln1+ln2+ln3+...+lnk\] \[e^{ln(x)}=e^{ln1+ln2+ln3+...+lnk}\] \[x=e^{ln1}e^{ln2}e^{ln3}...e^{lnk}\] \[x=1*2*3*...*k\]

OpenStudy (amistre64):

which ironically is: x = k!

OpenStudy (anonymous):

what do you mean is equal to k? do we multiply it?

OpenStudy (amistre64):

k is some arbitrary integer that we can count to; its real value is whatever number you stop at on the positive side of the number line

OpenStudy (amistre64):

|dw:1316703630934:dw|

OpenStudy (anonymous):

the answer in our book is 30.so it is just multiplied

OpenStudy (amistre64):

but to be fair, k can be any real number greater than 0 and it still amounts to the multiplication of all the stuff

OpenStudy (amistre64):

its just multiply the ln(innards) for that problem then yes

OpenStudy (amistre64):

ln(x) = ln(3) + ln(5) + ln(17) x = 3*5*17

OpenStudy (anonymous):

what if the given is like this 3 ln 2-ln 6+2 ln 3

OpenStudy (amistre64):

- means divide in log talk

OpenStudy (amistre64):

3 ln(2) would amount to ln(2^3) so just read it from left to right x = 2^3 / 6 * 3^2 ... and follow it in that order

OpenStudy (anonymous):

ah so the whole number before the ln becomes an exponent?did get I it right?

OpenStudy (anonymous):

ah so the whole number before the ln becomes an exponent?did get I it right?

OpenStudy (anonymous):

hey amistre64 are you still there?

OpenStudy (amistre64):

yes

OpenStudy (amistre64):

since logs are exponents; they act in the same way .... if that makes sense

OpenStudy (amistre64):

\[(B^4)^6=B^{4*6}=B^{24}\]

OpenStudy (anonymous):

ah so the whole number before the ln becomes an exponent?did get I it right?

OpenStudy (anonymous):

what does it mean?i don't understand

OpenStudy (amistre64):

\[log_B(4^6)=6\ lob_B(4)\]

OpenStudy (amistre64):

typoed it

OpenStudy (amistre64):

it means that 3 ln(2) = ln(2^3)

OpenStudy (anonymous):

ah so the whole number before the ln becomes an exponent?did get I it right?

OpenStudy (anonymous):

ah ok.i understand now.thanks a lot for the help...i'll try to answer another problem then to test if i really got it.thanks again

OpenStudy (amistre64):

youre welcome, good luck :)

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