Express the following in terms of log x and y \[log_3(\frac{x^2sqrt[3]{x-2}}{(2x+3)^5})\]
I got \[2log_3(x)+\frac{1}{3}log_3(x-2)-5log_3(2x+3)\] but is that what they asked for? in terms of x and y?
Write the question using equation editor..
Oh sorry..
@waterineyes \[\log_3(\frac{(x^2\sqrt[3]{x-2})}{(2x+3)^5})\]
You have very right till here but like you, I also did not get the question what it is asking about?
Where did you read this question can you attach that page here?
On a past final and no I am not able to... I don't have a scanner... but I double checked and those are the exact instructions and that is the question...
u got it right but what is y?!!
I don't know... I thought you guys may have seen these instructions before and recognize what to do...
So you need to use some properties of log Which law are you having trouble with?
\[\log_3(\frac{(x^2\sqrt[3]{x-2})}{(2x+3)^5}) \] You have division in there so why don't you first use \[\log_3(\frac{u}{v})=\log_3(u)-\log_3(v)\]
Is my answer directly at the top correct @myininaya?
It is correct if you are writing in terms of \[\log_3(x) \text{ and } \log_3(y)\] It seems though your directions say log(x) and log(y)
If so you have just a little more work
But what you have so far is correct! :)
So what do I do to get it to where it needs to be? :)
Recall the change of base formula: \[\log_3(u)=\frac{\ln(u)}{\ln(3)}=\frac{\ln(10)}{\ln(10)} \cdot \frac{\ln(u)}{\ln(3)}=\frac{\ln(10)}{\ln(3)} \frac{\ln(u)}{\ln(10)}\] \[=\log_3(10) \cdot \log_{10}(u) =\log_3(10)\log(u) \text{ note: }\log_{10}(u)=\log(u) \]
Okay...
Okay...I still don't know what to do..
Well I'm reading the instructions and it seems like it wants the parts containing the variables to be base 10
And I see no y Did you mean to write a y there?
What exactly do you instructions say?
Express the following in terms of logarithms of x and y
You know you have no y up there right?
Yes... there is no y on my question either that's why I am so confused...
But anyways it looks like you want the part containing the variable part to be base 10 that is why i was using change of base
Well, we didn't learn the change base formula at all so I guess it is not required for me to go any further from my answer... Thanks so much for all your help everyone!
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