Question

Can someone please help me ASAP. It's urgent!

Solve the following logarithmic equation.

2 log 3 - log x = 2

this is exactly how it's written, and i'd appreciate if someone explained the steps- not just told me the answer.

Answer 1

Umm, does it not give a base?
because then it isn't possible...

Answer 2

it doesn't really, but i believe the base is supposed to be 3?

Answer 3

so.... hmm

Answer 4

\(\bf 2 log(3) - log(x) = 2\) is that it?

Answer 5

yes i believe so! that's seriously how it's written though, not how it usually is with the bases very clear.

Answer 6

log(3^2 /x) = log2^2

Answer 7

if no log base is given, "base 10" is the given

Answer 8

i think it so

Answer 9

@jdoe0001 - when i have learned it log was without base log base 2

Answer 10

hmm have you got your log rules sheet around?

check how to get the "2log(3)" simplified first off

Answer 11

@jhonyy9 hmmm pretty sure is "10" when is just a bare "log"

Answer 12

but one never knows, maybe it was teh book, or just the teacher

Answer 13

i'm so confused, i believe 2log(3) is .954242509

Answer 14

but this i have learned log base ten just lg and without 10

Answer 15

yes, \[\log(x) = \log_{10}(x)\]by convention

Answer 16

@jakeenglish http://www.chilimath.com/algebra/advanced/log/images/rules%20of%20exponents.gif <--- use the 3rd rule there, for the 1st term
what does it give you?

Answer 17

so probably there in USA - but how i have learned it was log without base what mean log base 2

sorry

Answer 18

\(\bf {\color{brown}{ 2}}log(3)\qquad \qquad {\color{brown}{ a}}log(b)\iff log(b^{{\color{brown}{ a}}})\)

so...what would 2log(3) give you then?

Answer 19

notice, the rule 3 listed there

also notice the rule 2, we'll need that one shortly

Answer 20

sorry sorry i was working on another problem

Answer 21

yes?
no?
maybe?
no the foggiest?
was busy chasing the dog?

Answer 22

@jhonyy9 that's true in computer science, where most logs are binary, but in general, log(x) means either log base 10, or log base e (natural logarithm)

Answer 23

so it would be \[\log(3^2)\] ?

Answer 24

aha
yes.... so... let's us use the 2nd rule listed now
one that \(\bf log(3^2)-log(x)\qquad \qquad log_c(a)-log_c(b)\implies log_c\left( \frac{a}{b} \right)\)

so.. what would that give you, using the 2nd rule on that sheet?

Answer 25

give me one second, logs aren't my strong suit so sorry if this takes a little while for me to comprehend

Answer 26

k

Answer 27

i'm a little confused, because there's no base? or am i missing something?

Answer 28

@whpalmer4 - thank you - but i know what i have learned so lg mean log base ten and log natural was ln and not different way

Answer 29

ok the site crashed for me but i think i figured it out a little bit

\[\log(\frac{ 9 }{ x }) = 2\]

Answer 30

9 over 100 or .09

Answer 31

,hmm yes

Answer 32

hold the mayo

Answer 33

log 3^3 - log x = 2 ==> log (3^3/x) = 2 ==> 3^3/x = e^2. ==> x = 3^3/e^2. Or if it's base 10 then x = 3^3/10^2. just apply the log rules.

Answer 34

\(\bf log_{\color{red}{ a}}{\color{blue}{ b}}=y \iff {\color{red}{ a}}^y={\color{blue}{ b}}\qquad\qquad
% exponential notation 2nd form
{\color{red}{ a}}^y={\color{blue}{ b}}\iff log_{\color{red}{ a}}{\color{blue}{ b}}=y
\\ \quad \\ \quad \\
\textit{now, notice what you have}
\\ \quad \\
log\left( \cfrac{9}{x} \right)=2\implies log_{{\color{red}{ 10}}}\left( {\color{blue}{ \cfrac{9}{x} }} \right)=2\implies ?\)

what does that change to?

Answer 35

anyhow.. need to dash... so... ahemm
notice \(\bf log_{\color{red}{ a}}{\color{blue}{ b}}=y \iff {\color{red}{ a}}^y={\color{blue}{ b}}\qquad\qquad
% exponential notation 2nd form
{\color{red}{ a}}^y={\color{blue}{ b}}\iff log_{\color{red}{ a}}{\color{blue}{ b}}=y
\\ \quad \\ \quad \\
\textit{now, notice what you have}
\\ \quad \\
log\left( \cfrac{9}{x} \right)=2\implies log_{{\color{red}{ 10}}}\left( {\color{blue}{ \cfrac{9}{x} }} \right)=2\implies {\color{red}{ 10}}^2={\color{blue}{ \cfrac{9}{x} }}\)

pretty sure you can take it from there