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Solve the power equation $\;2\!\cdot\!3^{x^2}=6^x.$

Two equations

$$2 \cdot 3^{x^{2}} = 6^x $$ $$log_3(9) \cdot log_3(3^{x^{2}}) = log_3(6^x) $$ $$2x^2 - xlog_3(6) = 0 $$ $$x(2x - log_3(6)) = 0$$ $$x = 0$$ or $$x = \frac{log_3(6)}{2}$$ What do I do wrong? Task from school math about logarithm I has the first equation on the top of picture and try to logarithm them for different base. First is 3, second is 2 and third is 6 and everywhere get solution x = 0 but it is not a solution of first equation. Where I lost some limitations for the range of acceptable values or something else? I mean, I solve the second equation the same way and get right solutions I think...

  1. What do I do wrong?
  2. Is the easy way to check yourself solution with logarithm? Put the solution into equation when it comes logarithm is hard
  3. Where I lost some limitations?
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    $\begingroup$ Firefly, you should write $$\log_3\left(2\cdot3^{x^2}\right)=\log_36^x$$ $$\log_32+\log_33^{x^2}=\log_36^x$$instead of $$\color{red}{\log_39\,\cdot}\log_33^{x^2}=\log_36^x.$$ $\endgroup$
    – Angelo
    Commented Jul 6 at 13:48
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    $\begingroup$ Please don't use pictures of text or formulas. See here, why. Use MathJax. Here is a tutorial. $\endgroup$
    – Dietrich Burde
    Commented Jul 6 at 14:12
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    $\begingroup$ You can solve a quadratic too: $x^2\ln(3)-x\ln(6)+\ln(2)=0\iff (x-1)(\ln(3)x-\ln(2))=0$ $\endgroup$
    – Тyma Gaidash
    Commented Jul 6 at 14:23
  • $\begingroup$ @Angelo Could you write how it works? I think we make logarithm of each arguments but right is make logarithm of all part of equality $\endgroup$
    – Firefly
    Commented Jul 6 at 14:53
  • $\begingroup$ You have to use $log(a\cdot b)=log(a)+log(b)$ and you have to tell us how you come up with log(9) $\endgroup$
    – trula
    Commented Jul 6 at 15:33

1 Answer 1

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Put $2 = 3^a$ then $$2.3^{x^2} = 6^x \\ \iff 3^{a+x^2} = 3^x\cdot2^x \\ \iff 3^{a+x^2} = 3^x\cdot 3^{ax} \\ \iff x^2 - (a+1)x + a = 0 \\ \iff (x-1)(x-a) = 0 \\ \iff x =1,a$$

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