## Logarithms = ???

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### Logarithms = ???

So I'm taking AP BC Calc online at school, which is sort of hard to learn without a physical body there to ask questions and all that.

We're still in the phase of reviewing Trig/AB material and I'm already struggling - I dunno if I'm going to stick with the class, but anyways.

How would you go about solving this:

Solving for x.

I really have no idea how to go about doing this, any pointers would be awesome<3

We're still in the phase of reviewing Trig/AB material and I'm already struggling - I dunno if I'm going to stick with the class, but anyways.

How would you go about solving this:

Solving for x.

I really have no idea how to go about doing this, any pointers would be awesome<3

log(4)=.60206

log(2)=.30103

I'm terribly rusty with this whole process but hopefully it'll all come back soon enough.

Edit:

Just kidding: I just actually looked at the decimals and realized log(4)/log(2) = 2. Lemme try this again.

Edit #2:

Ok, I got x=-2, but when I find the intercept with my calculator, it's at x=4.

Here's my work:

Edit #3:

I think I got it. Instead of making it 2log(9x) I left it as log(9x^2) and turned the other side into log(5x-8 )^2. Cancel the log's, square root both sides to get 3x=5x-8, simplify, x=4.

Also: I appreciate the help

This is getting easier and easier.

I'm on to expanding/compressing log/ln's now and was wondering if there's any way to expand ln(x^3 - 4) any more than it already is.

I don't think there is, because as far as I know it's only considered expanding if there are two factors being multiplied/divided.

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### Re: Logarithms = ???

You don't always need to change to base e. This problem's simpler if you don't:

Code: Select all

`Simplify LHS:`

log_4(9x^2) = log_2 ((3x)^2) / log_2 (4)

= (1/2)*2 log_2 (3x)

Then log_2 (3x) = log_2 (5x-8)

3x=5x-8

x=4

### Re: Logarithms = ???

Strychnos wrote:

Solving for x.

log_4(9 x^2) = log_2(5x - 8)

log_4(9 x^2) = log_2(9 x^2) * log_4(2) = log_2(9 x^2) * 1/2

log_2(9 x^2) = 2 * log_2(5x - 8) = log_2( (5x - 8)^2 )

(3x)^2 = (5x-8)^2

3x = (+/-) (5x - 8)

x = 1 or 4.

I think.

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Pesto wrote:Looking at the cheat sheet there is an identity for addition. I hadn't seen it before, but that's not saying much.

With the addition identity, they just factored out 'a' from inside the logarithm, then used the division identity on the 'c/a' left over in the log(1 +/- c/a) logarithm, and then for some expressed it as an exponential with base b...

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log4 9x^2 = log2 5x-8 is the problem

with problems like this, the easiest thing to check for first is if you can get both sides in the same base. In this case you can - taking the square root of both the base and argument of a log will not change the number (neither will taking the cube root, or squaring it, etc). So, taking the square root of the left, we get

log2 3x = log2 5x-8

now if logx a = logx b, then a = b. So;

3x = 5x-8

Let's hope you can solve that, if you cant it's my opinion that you should drop calc BC

Just think of what a logarithm represents.

So:

log4 9x^2 = log2 5x-8

9x^2 = 4^(log2 5x-8)

because

Notice that the only remaining base is 2, and 4 is a power of 2.

So since 4 = 2^2, we have

9x^2 = (2^2)^(log2 5x-8)

and by rules for exponents, you know it's the same as

9x^2 = 2^(2*(log2 5x-8))

Now a good way to solve this problem might be to take a square root of both sides to get 3x = 2^(log2 5x-8). But that's no fun.

What is fun is the following:

9x^2 = 2^(log2 ((5x-8)^2))

Now 2^log2(something inside) = something inside, since

So that gets us to:

9x^2 = (5x-8)^2

Now it's profitable to note that 9 is a perfect square--that is, 3^2 = 9.

So (3x)^2 = (5x-8)^2

So 3x = 5x-8,

2x = 8,

x = 4.

So:

log4 9x^2 = log2 5x-8

9x^2 = 4^(log2 5x-8)

because

Highlight for the key to all log problems wrote:Log is the inverse of an exponential.

Notice that the only remaining base is 2, and 4 is a power of 2.

So since 4 = 2^2, we have

9x^2 = (2^2)^(log2 5x-8)

and by rules for exponents, you know it's the same as

9x^2 = 2^(2*(log2 5x-8))

Now a good way to solve this problem might be to take a square root of both sides to get 3x = 2^(log2 5x-8). But that's no fun.

What is fun is the following:

9x^2 = 2^(log2 ((5x-8)^2))

Now 2^log2(something inside) = something inside, since

Highlight for the key to all log problems wrote:Log is the inverse of an exponential.

So that gets us to:

9x^2 = (5x-8)^2

Now it's profitable to note that 9 is a perfect square--that is, 3^2 = 9.

So (3x)^2 = (5x-8)^2

So 3x = 5x-8,

2x = 8,

x = 4.

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Code: Select all

`log4(9x^2) = log2(5x-8)`

log2(sqrt(9x^2)) = log2(3x) = log2(5x-8)

3x = 5x-8

2x = 8

x = 4

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