The problem is to express $4\ln(x)+2\ln(x^4y^3)+5\ln(z)$ as a single logarithm.
Our teacher has shown us examples for the same base and when it's both add and subtract. But I'm not sure how to do this.
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ Can you elaborate on your title? Does the final result need to be in log base-3? $\endgroup$– Kaj HansenCommented Apr 29, 2014 at 0:51
-
$\begingroup$ I think Blake just wants to: "express the sum of three logarithms as a single logarithm," but the title is a little unclear. $\endgroup$– Graham KempCommented Apr 29, 2014 at 0:54
-
$\begingroup$ Sorry, I was only wanting to express this as a single logarithm. :) $\endgroup$– BlakeCommented Apr 29, 2014 at 0:57
-
$\begingroup$ Welcome to math stack exchange!! $\endgroup$– user122519Commented Apr 29, 2014 at 4:34
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
7
Hint: $r\ln x=\ln(x^r)$, so $4\ln(x)=\ln(x^4)$. Also, $\ln(x)+\ln(y)+\ln(z)=\ln(xyz)$.
-
$\begingroup$ Good answer (+1), but it's unclear from the OP whether he needs his final answer in $\log_3$. $\endgroup$ Commented Apr 29, 2014 at 0:50
-
$\begingroup$ @KajHansen Hmm, yeah, that's true. I read the title and interpreted it as log base $3$. If not, then the last sentence is redundant. $\endgroup$– user122283Commented Apr 29, 2014 at 0:53
-
$\begingroup$ Sorry, I do not need my answer in log3. I've gotten to ln(x^4)+ln(x^4y^3)^2+ln(z)^5. $\endgroup$– BlakeCommented Apr 29, 2014 at 0:53
-
$\begingroup$ @Blake So I'll edit my answer. Please read it again in five seconds. $\endgroup$– user122283Commented Apr 29, 2014 at 0:54
-
$\begingroup$ I'm not 100% confident in this answer, but I got ln(x^16y^3z^5)^10. The ^10 at the end is making me unsure of the process I'm using. $\endgroup$– BlakeCommented Apr 29, 2014 at 0:56
$\begingroup$
$\endgroup$
Hints:
$\log(x) + \log(y) = \log(xy)$ and $n\log(x) = log(x^n)$ for logarithms of any base.
answered Apr 29, 2014 at 0:44
$\begingroup$
$\endgroup$
1
Are you familiar with the identity $r \ln x = \ln (x^r)$?
I don't believe you need to change the base of any logarithms here, by the way.
-
$\begingroup$ Oops, I think I might've misinterpreted his question. I thought the final result had to be in $\log_3$ due to the title. $\endgroup$ Commented Apr 29, 2014 at 0:47
$\begingroup$
$\endgroup$
$4\ln(x)+2\ln(x^4y^3)+5\ln(z)$
By using identity, $a\times ln x=ln x^a$
$\implies \ln(x)^4+\ln{(x^4y^3)}^2+\ln(z)^5$
By using identity, $\ln x+ \ln y=\ln {xy}$
$\implies \ln(x)^{4+8}y^{3\times2}(z)^5$
Final Answer,
$\implies \ln(x)^{12}y^{6}(z)^5$
