# Math Help - 3 math problems.

1. ## 3 math problems.

Hi...
I tried to solve them, but ended with nothing :-(

Thank you!!!

2. Originally Posted by rectangle
Hi...
I tried to solve them, but ended with nothing :-(

Thank you!!!

you realize that

$5^4 = 625$

so you have

$625 ^x = 5^{4x}$ and $25^x = 5^{2x}$ and $125^x = 5^{3x}$

does this help you in any way ? show what have you try to do

3. Originally Posted by rectangle
Hi...
I tried to solve them, but ended with nothing :-(

...

Thank you!!!
To #1:

1. Probably you have noticed that $5^4 =625$ , $5^3=125$, $5^2 = 25$

$5^{4x}+3+6\cdot 5^{2x} = -4\cdot 5^x+2-4\cot 5^{3x}$

3. Rearrange the equation:

$5^{4x}+4\cot 5^{3x}+6\cdot 5^{2x} +4\cdot 5^x+1= 0$

4. This is a complete polynom of 4th degree:

$(5^x+1)^4=0$

This equation doesn't have any real solution.

4. Hello, rectangle!

$\text{Solve: }\;81^x - 12\cdot27^x + 47\cdot9^x \;=\;72\cdot3^x - 36$

We have: . $(3^4)^x - 12(3^3)^x + 47(3^2)^x - 72(3)^x + 36 \;=\;0$

. . . . . . . . $(3^x)^4 - 12(3^x)^3 + 47(3^x)^2 - 72(3^x) + 36 \;=\;0$

Let $u \,=\,3^x\!:\;\;u^4 - 12u^3 + 47u^2 - 72u + 36 \;=\;0$

Factor: . $(u-1)(u-2)(u-3)(u-6) \;=\;0$

. . Hence: . $u\;=\;1,\:2,\:3,\:6$

Back-substitute:

. . $\begin{array}{ccccccc}
3^x \;=\; 1 & \Rightarrow & x \;=\; 0 \\
3^x \;=\; 2 & \Rightarrow & x \;=\; \log_32 \\
3^x \;=\; 3 & \Rightarrow & x \;=\; 1 \\
3^x \;=\; 6 & \Rightarrow & x \;=\; \log_36 \end{array}$

5. Originally Posted by rectangle
Hi...
I tried to solve them, but ended with nothing :-(

...

Thank you!!!
To #3.1:

$2\left(\log_9(x)\right)^2=1+\log_9(x)$

Use the substitution $u = \log_9(x)$. Your equation becomes:

$2u^2=1+u$. Solve for u.

Afterwards re-substitute and solve for x. I've got $x = 9 ~\vee~x=\frac13$

6. Originally Posted by yeKciM
you realize that

$5^4 = 625$

so you have

$625 ^x = 5^{4x}$ and $25^x = 5^{2x}$ and $125^x = 5^{3x}$

does this help you in any way ? show what have you try to do
Thank you. It did help me a lot!

Originally Posted by earboth
To #1:

1. Probably you have noticed that $5^4 =625$ , $5^3=125$, $5^2 = 25$

$5^{4x}+3+6\cdot 5^{2x} = -4\cdot 5^x+2-4\cot 5^{3x}$

3. Rearrange the equation:

$5^{4x}+4\cot 5^{3x}+6\cdot 5^{2x} +4\cdot 5^x+1= 0$

4. This is a complete polynom of 4th degree:

$(5^x+1)^4=0$

This equation doesn't have any real solution.
Thank you so much.

Originally Posted by earboth
To #3.1:

$2\left(\log_9(x)\right)^2=1+\log_9(x)$

Use the substitution $u = \log_9(x)$. Your equation becomes:

$2u^2=1+u$. Solve for u.

Afterwards re-substitute and solve for x. I've got $x = 9 ~\vee~x=\frac13$
Thank you. Can I solve the second one (3.2) also in that way?

Originally Posted by Soroban
Hello, rectangle!

We have: . $(3^4)^x - 12(3^3)^x + 47(3^2)^x - 72(3)^x + 36 \;=\;0$

. . . . . . . . $(3^x)^4 - 12(3^x)^3 + 47(3^x)^2 - 72(3^x) + 36 \;=\;0$

Let $u \,=\,3^x\!:\;\;u^4 - 12u^3 + 47u^2 - 72u + 36 \;=\;0$

Factor: . $(u-1)(u-2)(u-3)(u-6) \;=\;0$

. . Hence: . $u\;=\;1,\:2,\:3,\:6$

Back-substitute:

. . $\begin{array}{ccccccc}
3^x \;=\; 1 & \Rightarrow & x \;=\; 0 \\
3^x \;=\; 2 & \Rightarrow & x \;=\; \log_32 \\
3^x \;=\; 3 & \Rightarrow & x \;=\; 1 \\
3^x \;=\; 6 & \Rightarrow & x \;=\; \log_36 \end{array}$

Thank you. I did not understand the "Factor" Part. It would be great if someone could explain me in more details.

Thank you all
I feel like an idiot.