9^lnx-4*3^lnx+1 + 27=0

Question

zehanz · Answer

Remember 9=3², so the formula can be rewritten as:

$3^{2^{\ln x}}-4\cdot3^{\ln x}+27=0$

and then, because $3^{2^{\ln x}}=3^{2{\ln x}}=(3^{\ln x})^2$:

$(3^{\ln x})^2-4\cdot3^{\ln x}+27=0$.

Doesn't that remind you of quadratic equations (set $p=3^{\ln x}$)?

zehanz · Answer

I used the rule: $(a^{b})^c=a^{bc}=a^{cb}=(a^{c})^b$

anonymous · Answer

3^2lnx−4⋅3^lnx+1+27=0      ...you did not put +1 in force ..is the same thing withaout it??

zehanz · Answer

Not quite :(
I overlooked that 1.
Should it be ...^ln(x+1)+27 =...  ? Otherwise it would just be ...^ln(x) + 28 = ...

zehanz · Answer

Oh, I see now what it has to be:

$(3^{\ln (x)})^2-4 \cdot 3^{\ln(x) +1}+27=0$, so this is
$(3^{\ln (x)})^2-4 \cdot 3^{\ln(x)}\cdot 3^1+27=0$
$(3^{\ln (x)})^2-12 \cdot 3^{\ln(x)}+27=0$
Now you can factor it into

$(3^{\ln x}-9)\cdot(3^{\ln x}-3)=0$, which gives

$3^{\ln x}=9$ or $3^{\ln x}=3$, meaning

$\ln x=2$ or $\ln x=1$, so

$x=e^2$ or $x=e$