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).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
Hey having a lot of trouble with this problem.
Please show all steps
Assuming 3 and 9 are "bases of logarithm"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
BBB, while replying to a post, please use the "reply" button of the post - then the OP cannot be erased.log3(x−4)=1+log9(x)log(3)log(x−4)=1+log(9)log(x)log(3)log(x−4)−1−log(9)log(x)=0Combine into one fraction and continue...
Noted. ?BBB, while replying to a post, please use the "reply" button of the post - then the OP cannot be erased.
I found this one particularly helpful. So thank you!!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