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how about if the log4 were to be log45. What would the answer be?
Thanks.
I had 20.
The book has 10 as the answer
Sorry I supposed to have checked the writings before posting.
Thanks.
Please check, the following for me.
Log(2x+3)=log4.
I had 20 as the answer, the has 10.
If 2logy-log2x=2log(y-x), express y in terms of x. I had (y-x)(y-2x^2). The book has y^-4xy+4^2x=0.
log(10+9x)-log(11-x)=2. I had 20 as answer, the book has 10.
Last but not the least
log81/log3^1.
the book has -4 at its back as the answer.
Thanks.
Please, check the above if the answers are right. Me vs the book. -:)
Hi;
Factor out a 3^(x-1) from each term on the left.
I don't understand how you factored it out.
I have considered it carefully, but see no way to solve than what I have done, please assist.
Thanks.
The final result is my answer
Okay, I admit, then see my thoughful manipulation:
3^x + 3^x-1= 4
multiplying 3 through.
3*3^x + 3^x*1/3 = 4* 3.
Let 3^x=m. therefore,
=3m + m = 12
= 4m - 12 =
0
factorising.
= 4(m-3) =0.
Is this correct?
How then would one solve this?
The book has x=0 or 1 as answer at its back.
I had zero as the answer.
But as I am looking through the book, I chanced on the following, which I have solved but one of the answers was impossible to get. And the book had zero with that answer.
3^x + 3^-1=4.
My final answers are, x=1, and the impossible answer, 3^x=4.
Please, see how I solved the following:
2^x +2^-x=2.
bases are the same, therefore equate the exponents.
x-1-x =1
= x-x-1=1
=0-1 =1
=0 = 1+1
=0=2.
Is that right?
I think bases are the same, so doesnt need logarithm application.
You had your final answer by this:
x = log2/log2+log3
= 1.477
But I cant tell why the has, 0.39
Okay, but bobbym, the law logx-logy= logx/logy.
If it had been, logy-logy=0, is the zero?
If the logs have the same figures, it not right for one to divide?
But rather subtraction must take place?
Thank you very much bobbym for the link.
But there was xlog3 also present at the left side with the log2. Why didn't you balance the eqution with the log3 instead? because it was also present at the left side.
I just want to why you didnt use the log3 instead.
Thanks.
But I am sure if you had not added log2 to both sides you would still arrive at the same answer, is not that?
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xlog2 - "log2" +xlog3 =0. From this stage you took off the "log2", why that happened.
Okay, I will provide the steps:
Take the log of both sides.
Add log(2) to both sides.
.
Then how did you arrived at the last stage? Could you explain to me?
Why did you added log two to both sides?
Thanks
This is what I mean, and it is one of the laws of logarithm in my book:
logx - logy = log(x/y).
This is what I mean, and I see you did not apply it at that step. Why didn't you?
Okay, I will provide the steps:
Take the log of both sides.
Why the last step is not:
Since the law says. for instance log2-log2 = log2/log2.
Please provide a link for that.
Is not among the law of log in my book.
And why log10^2 in # 163 correct?
Please I what you to answer my question at #161
Thats what is written in the book.
It used log ten for the unknown log of the two. Hence
2-2log5=log10^2-log5^2
But I dont know why.
Is that wrong if so how should it be?
How different is this x-1log2 in the previous problem from 2-2log5 in the example.
Because I see it that one must apply the same method, but the method used to solve both are different.
Please see this;
x-1 log2 = xlog2-1log2.
(The same base used).
And
2-2log5 = log10^2-log5^2.
(Different bases used).
why is it log10^2 and not log5^2 instead. So that the whole thing would be;
log5^2-log5^2,
Please, explain to me why.
I am comparing two things because I dont understand your step 2.
The is an example from the book: 2-2log5 = log10^2 - log5^2 = log4
so does 2-2log5 mean log5^2-2?
That what I want to know.
The book has this example.
2 - 2log5 = log10 - log5^2 =log4
why should not it be, 2 - 2log5 = "log5^2" -log5^2 so that it will go by the law.
With example it has log10^2, I think should be log5^2
Yes, I understand that law, so now, log2^x-1= xlog2 - log2
So does my example mean, 2 - 2log5 = log5^2-2.
If so why is it equall to log10^2 - log5^2. and not log5^2 - log5^2 in order to go according to the law.