Log help: ???3(? − 4) = 1 + ???9(?)

[sophierlz]

Hey having a lot of trouble with this problem. I don't know how to get an answer at all

----> log base 3 (x-4) = 1 + log base 9 (x)

???3(? − 4) = 1 + ???9(?)


Please show all steps

Thank you xx

 

[BigBeachBanana]

[math]\log_3(x-4) = 1 + \log_9(x)\\ \frac{\log(x-4)}{\log(3)}= 1 + \frac{\log(x)}{\log(9)}\\ \frac{\log(x-4)}{\log(3)}- 1 -\frac{\log(x)}{\log(9)}=0\\[/math]Combine into one fraction and continue...

 

[Dr.Peterson]

Hey having a lot of trouble with this problem. I don't know how to get an answer at all

----> log base 3 (x-4) = 1 + log base 9 (x)

???₃(? − 4) = 1 + ???₉(?)


Please show all steps

Thank you xx

Since 9 is a power of 3, you might try using the change-of-base formula, as BBB did, but using base 3 as the "new" base. This will make some of the work easier. Or, alternatively, express log(9) in terms of log(3).

Then try standard methods you will have learned for logarithmic equations: rewrite ("condense") as a single log.

 

[lookagain]

Hey having a lot of trouble with this problem.

Please show all steps

We are not here to show you "all steps." When you post
a problem, post what you know with it so that we
may help you with where you are stuck.

 

[Deleted member 4993]

Hey having a lot of trouble with this problem. I don't know how to get an answer at all

----> log base 3 (x-4) = 1 + log base 9 (x)

???3(? − 4) = 1 + ???9(?)


Please show all steps

Thank you xx

Assuming 3 and 9 are "bases of logarithm"

???3(? − 4) = 1 + ???9(?) ...............convert log₉(t) to log₃(t)

???₃(? − 4) = 1 + (1/2) *???₃(?)

???₃(? − 4) - (1/2) *???₃(?) = 1

???₃(? − 4) - ???₃(√?) = 1..................................... now continue

 

[Harry_the_cat]

Come on, Sophierlz. Let us know if this has helped? Tell us what bit you don't understand.

Or , if it has helped, please share your solution for others.

 

[Deleted member 4993]

[math]\log_3(x-4) = 1 + \log_9(x)\\ \frac{\log(x-4)}{\log(3)}= 1 + \frac{\log(x)}{\log(9)}\\ \frac{\log(x-4)}{\log(3)}- 1 -\frac{\log(x)}{\log(9)}=0\\[/math]Combine into one fraction and continue...

BBB, while replying to a post, please use the "reply" button of the post - then the OP cannot be erased.

 

[BigBeachBanana]

BBB, while replying to a post, please use the "reply" button of the post - then the OP cannot be erased.

Noted. ?

 

[sophierlz]

yes i figured it out now, thanks to everyone;

Assuming 3 and 9 are "bases of logarithm"

???3(? − 4) = 1 + ???9(?) ...............convert log₉(t) to log₃(t)

???₃(? − 4) = 1 + (1/2) *???₃(?)

???₃(? − 4) - (1/2) *???₃(?) = 1

???₃(? − 4) - ???₃(√?) = 1..................................... now continue

I found this one particularly helpful. So thank you!!
Also Big Beach Banana was a huge help.!!
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