How much do you know about logarithms? To me these questions appear to be elementary applications of the laws of logs. Could you be more specific as to where you're having problems?
Here are some ln (lawn) and e^x questions. I'm still learning these so I'm not so good at solving thema and these ones are hard so I was wondering if anyone can help.
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Hello, SportfreundeKeaneKent!
Here are a couple of them . . .
2) Solve: .
We have: .
Factor: .
And we have two equations to solve:
. .
. .
3) The graph of is rotated 90° CCW about the origin.
What is the equation of the new graph?
The graph of looks like this:Code:| | * | * ------+-------*---------- | * 1 | * | * | |* |
Rotated 90° CCW, it looks like this:Code:* | | * | * | * | * | * | * | * ------------+-------------------- |
We're expected to recognize this as the graph of:
We're sort of doing the log unit at school right now. I thought that these seemed pretty simple too but there's always one extra thing you have to do on some of these harder problems that the teacher gives. The ones from the textbook aren't a problem. lnx is a bit more of a problem and I'm still learning that.
Oh and also, I was doing question 4 down that list and I got x = 0.5 which is different than the answer Soroban's answer of 1/e. I was wondering if anyone could check up on this.
I used the quotient rule in this question so that it became this:
ln(e^2 x^2+1)-ln2x=1
2lne^2 x+ 0 - ln2x=1
2x + 0 - ln2x = 1
2x(1-ln1)=1
2x=1
x-1/2 or 0.5
When I plugged x = 0.5 into the equation to check my work, it worked out so I'm really confused about wheter what Soroban did was right or the way I've done it is right.
Hello Sportsfreund,
here:
ln(e^2 x^2+1)-ln2x=1
2lne^2 x+ 0 - ln2x=1
you've made a very common mistake: You used the "ln" as a kind of variable and expanded the brackets by rule of distribution. But ln is a function and the contents of the brackets are the arguments of this function.
You can get rid of the "ln" if you use the following property:
Soroban's solution is 100% correct.
EB