To #1:
1. Probably you have noticed that $\displaystyle 5^4 =625$ , $\displaystyle 5^3=125$, $\displaystyle 5^2 = 25$
2. Your equation becomes:
$\displaystyle 5^{4x}+3+6\cdot 5^{2x} = -4\cdot 5^x+2-4\cot 5^{3x}$
3. Rearrange the equation:
$\displaystyle 5^{4x}+4\cot 5^{3x}+6\cdot 5^{2x} +4\cdot 5^x+1= 0$
4. This is a complete polynom of 4th degree:
$\displaystyle (5^x+1)^4=0$
This equation doesn't have any real solution.
Hello, rectangle!
$\displaystyle \text{Solve: }\;81^x - 12\cdot27^x + 47\cdot9^x \;=\;72\cdot3^x - 36$
We have: .$\displaystyle (3^4)^x - 12(3^3)^x + 47(3^2)^x - 72(3)^x + 36 \;=\;0 $
. . . . . . . .$\displaystyle (3^x)^4 - 12(3^x)^3 + 47(3^x)^2 - 72(3^x) + 36 \;=\;0$
Let $\displaystyle u \,=\,3^x\!:\;\;u^4 - 12u^3 + 47u^2 - 72u + 36 \;=\;0$
Factor: . $\displaystyle (u-1)(u-2)(u-3)(u-6) \;=\;0 $
. . Hence: .$\displaystyle u\;=\;1,\:2,\:3,\:6$
Back-substitute:
. . $\displaystyle \begin{array}{ccccccc}
3^x \;=\; 1 & \Rightarrow & x \;=\; 0 \\
3^x \;=\; 2 & \Rightarrow & x \;=\; \log_32 \\
3^x \;=\; 3 & \Rightarrow & x \;=\; 1 \\
3^x \;=\; 6 & \Rightarrow & x \;=\; \log_36 \end{array}$
Thank you. It did help me a lot!
Thank you so much.
Thank you. Can I solve the second one (3.2) also in that way?
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.