# Math Help - More log stuff

1. ## More log stuff

1. $\log_{2}(25^(x+3) - 1) = 2 + \log_{2}(5 ^ (x+3) + 1)$
2. $\log_{4}(2 \cdot 3^(x) - 6) -\log_{2}(9^(x) - 6)$
3. $\log_{4}(2 \cdot 4^(x-2) - 1) + 4 = 2x$

I attempted the first to come as far as this-:
Taking x+3 = n,
$25^(n) - 1/5^(n) + 1 = 4$
I've forgotten exponential mathematics it seems..
Pardon me if I seem like a total noob, but I need help :P

2. Hello,

For latex use, do x^{...}, not x^(...)

Originally Posted by SVXX
1. $\log_{2}(25^(x+3) - 1) = 2 + \log_{2}(5 ^ (x+3) + 1)$
2. $\log_{4}(2 \cdot 3^(x) - 6) -\log_{2}(9^(x) - 6)$
3. $\log_{4}(2 \cdot 4^(x-2) - 1) + 4 = 2x$

I attempted the first to come as far as this-:
Taking x+3 = n,
$25^(n) - 1/5^(n) + 1 = 4$
I've forgotten exponential mathematics it seems..
Pardon me if I seem like a total noob, but I need help :P
Be careful with logs : $\log (a+b) \neq \log a+ \log b$

$\log_{2}(25^n - 1) = 2 + \log_{2}(5^n + 1)$

$\log_2 (25^n -1)-\log_2 (5^n+1)=2$

Using the identity : $\log a-\log b=\log \frac ab$ :

$\log_2 \frac{25^n-1}{5^n+1}=2$

$\frac{25^n-1}{5^n+1}=4$

$25^n-1=4 \cdot 5^n+4$

Substitute $u=5^n$

3. I still didn't get it!
$25^{n} - 4\cdot u = 5$
$5^{2n} - 4\cdot u = 5$
Have to get rid of that 2n somehow!

4. Originally Posted by SVXX
I still didn't get it!
$25^{n} - 4\cdot u = 5$
$5^{2n} - 4\cdot u = 5$
Have to get rid of that 2n somehow!
$5^{2n}=(5^n)^2=u^2$

5. Alright, now I know for sure my head is sleeping.
From that, x + 3 = 1, x = -2.
Skipping the 2nd one, there's no equation to it...
In the 3rd,
$\log_{4}(2 \cdot 4^{x-2} - 1) + \log_{4}256 = 2 \cdot x$
$\log_{4}((2 \cdot 4^{x-2} -1) \cdot 256) = 2 \cdot x$
$((2 \cdot 4^{x-2} - 1) \cdot 256) = 4^{2x}$
Some help on this? Your advice comes really quick, thanks a lot for that

6. Originally Posted by SVXX
Alright, now I know for sure my head is sleeping.
From that, x + 3 = 1, x = -2.

Skipping the 2nd one, there's no equation to it...
Yep, strange

In the 3rd,
$\log_{4}(2 \cdot 4^{x-2} - 1) + \log_{4}256 = 2 \cdot x$
$\log_{4}((2 \cdot 4^{x-2} -1) \cdot 256) = 2 \cdot x$
$((2 \cdot 4^{x-2} - 1) \cdot 256) = 4^{2x}$
Some help on this?
Transform the powers..

$4^{x-2}=2^{2x-4}$

$4^{2x}=2^{4x}=(2^{2x})^2$

--> $(2^{2x-3}-1) \cdot 2^8=(2^{2x})^2$

$2^{2x+5}-2^8=(2^{2x})^2$

$2^5 \cdot 2^{2x}-2^8=(2^{2x})^2$

Is there a substitution you can make ?

Your advice comes really quick, thanks a lot for that
haha, you're welcome =)

7. Originally Posted by Moo

Yep, strange

Transform the powers..

$4^{x-2}=2^{2x-4}$

$4^{2x}=2^{4x}=(2^{2x})^2$

--> $(2^{2x-3}-1) \cdot 2^8=(2^{2x})^2$

$2^{2x+5}-2^8=(2^{2x})^2$

$2^5 \cdot 2^{2x}-2^8=(2^{2x})^2$

Is there a substitution you can make ?

haha, you're welcome =)
Oh yeez, those substitutions
$2^{5} \cdot 2^{2x} - 2^{8} = (2 \cdot 2x)^2$
Taking 2 raised to 2x as u,
$u^{2} - 32 \cdot u + 256 = 0$
We get equal roots, u = 16.
$2^{2x} = 16$
$2x = 4$
$x = 2$
Thanks for the support!
And don't mind if I come up with a newby sum again :P
Vous êtes les meilleurs! ^_^

8. Originally Posted by SVXX
Thanks for the support!
And don't mind if I come up with a newby sum again :P
Vous êtes les meilleurs! ^_^
It's not really newby oO
All of this comes with training ^^

Haha, good French speaking ^'

9. I'm not so good at French, but still..:P
Ok, the last problem. And this one isn't so sweet as the others were...

$|\log_{2}(\frac{4-x}{2x - 5})| + \log_{2}x = 1$
There's a modulus sign...there are separate ways of solving modulus problems..I guess I have to integrate those ways.
But yeah, better off is to ask you =P

10. Originally Posted by SVXX
I'm not so good at French, but still..:P
Ok, the last problem. And this one isn't so sweet as the others were...

$|\log_{2}(\frac{4-x}{2x - 5})| + \log_{2}x = 1$
There's a modulus sign...there are separate ways of solving modulus problems..I guess I have to integrate those ways.
But yeah, better off is to ask you =P

Differentiate the case when $\frac{4-x}{2x-5}>1$ (then $|\log_{2}(\frac{4-x}{2x - 5})|=\log_{2}(\frac{4-x}{2x - 5})$) and when $\frac{4-x}{2x-5}<1$ (then $|\log_{2}(\frac{4-x}{2x - 5})|=-\log_{2}(\frac{4-x}{2x - 5})$)

I think it's quite tricky... Because you have to find the domain when it's <1 and when it's >1...

It's the only way I see so far...

11. Originally Posted by Moo

Differentiate the case when $\frac{4-x}{2x-5}>1$ (then $|\log_{2}(\frac{4-x}{2x - 5})|=\log_{2}(\frac{4-x}{2x - 5})$) and when $\frac{4-x}{2x-5}<1$ (then $|\log_{2}(\frac{4-x}{2x - 5})|=-\log_{2}(\frac{4-x}{2x - 5})$)

I think it's quite tricky... Because you have to find the domain when it's <1 and when it's >1...

It's the only way I see so far...
Yeah, there will be two cases here..
If it is > 1, then
$x = \sqrt{10}$
If it is < 1, then
$x = 2.88, -1.38$
$x = -1.38$
Thats how it came via calculations...correct me if I'm wrong!

12. Originally Posted by SVXX
Yeah, there will be two cases here..
If it is > 1, then
$x = \sqrt{10}$
(Or $-\sqrt{10}$)

It's $\frac{4-x}{2x-5}>1$, not x>1

But after you've found the roots, you have to check if it's really >1.

$\frac{4-\sqrt{10}}{2x-5} \approx 0.632 <1$

So $\sqrt{10}$ is not a solution.

This goes the same way for $-\sqrt{10}$

If it is < 1, then
$x = 2.88, -1.38$
$x = -1.38$
Once again, don't check if x<1, but if $\frac{4-x}{2x-5}<1$

13. Originally Posted by Moo
(Or $-\sqrt{10}$)

It's $\frac{4-x}{2x-5}>1$, not x>1

But after you've found the roots, you have to check if it's really >1.

$\frac{4-\sqrt{10}}{2x-5} \approx 0.632 <1$

So $\sqrt{10}$ is not a solution.

This goes the same way for $-\sqrt{10}$

Once again, don't check if x<1, but if $\frac{4-x}{2x-5}<1$
In that case, x = -1.38!
You're a femme fatale when it comes to maths

14. Originally Posted by SVXX
In that case, x = -1.38!
Yes

But the most important thing : do you understand why ?

You're a femme fatale when it comes to maths

15. Originally Posted by Moo
Yes

But the most important thing : do you understand why ?

Oh yez I know :P
Next time I'll ask you directly when I get a problem lol

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