Exponents with different bases

[Mentallic]

Homework Statement

This is re-asking the question in https://www.physicsforums.com/showthread.php?t=273275" which mysteriously went quiet before the answer could be given... Anyway I also tried answering the question, but to no avail.

Homework Equations

[tex]a^m.a^n=a^{m+n}[/tex]
[tex](a^m)^n=a^{mn}[/tex]
[tex]log(ab)=loga+logb[/tex]
[tex]log(a^b)=bloga[/tex]

The Attempt at a Solution

[tex]12^x=4.8^{2x}[/tex]
[tex]3^x.(2^2)^x=2^2(2^3)^{2x}[/tex]
[tex]3^x.2^{2x}-2^{6x+2}=0[/tex]
[tex]2^{2x}(3^x-2^{4x-2})=0[/tex]

Hence, [tex]2^{2x}=0[/tex] (1) or [tex]3^x-2^{4x-2}=0[/tex] (2)
(1) has no solution.

(2) [tex]\Rightarrow[/tex] [tex]3^x=2^{4x+2}[/tex]
[tex]xlog3=(4x+2)log2[/tex]

How can one go about solving this? different bases? It gives an answer of [tex]x\approx -0.82814[/tex]

 

[Mentallic]

Oh hang on...

[tex]xlog3=(4x+2)log2[/tex]
[tex]xlog3=4xlog2+2log2[/tex]
[tex]x(log3-4log2)=2log2[/tex]
[tex]x=\frac{2log2}{log3-4log2}[/tex]

[tex]x\approx -0.82814449[/tex]

Ahh never mind.

 

[rosh300]

12^x = 4X8^(2x)
take logs of both sides: log(12^x) = log(4 X 8^2x)
Get rid of the power 2 (althought not essential) log(12^x) = log(4 X 64^x)
Split RHS using the laws of logs and bring the powers out: x log(12) = log(4) + x log(64)
see if u can do the last step (hint collect all terms with x)

 

[Дьявол]

It would rather use log₃.