Respuesta :

[tex]\\ \rm\Rrightarrow 36^x-7(6^x)-18=0[/tex]

[tex]\\ \rm\Rrightarrow 6^{2x}-7(6^x)-18=0[/tex]

[tex]\\ \rm\Rrightarrow (6^x)^2-7(6^x)-18=0[/tex]

  • Let 6^x=p

[tex]\\ \rm\Rrightarrow p²-7p-18=0[/tex]

[tex]\\ \rm\Rrightarrow p²-9p+2p-18=0[/tex]

[tex]\\ \rm\Rrightarrow p(p-9)+2(p-9)=0[/tex]

[tex]\\ \rm\Rrightarrow (p+2)(p-9)=0[/tex]

Omit negative as we have to solve for real nos

  • p=9

put value

[tex]\\ \rm\Rrightarrow 6^x=9[/tex]

[tex]\\ \rm\Rrightarrow ln6^x=ln3^2[/tex]

[tex]\\ \rm\Rrightarrow xln6=2ln3[/tex]

[tex]\\ \rm\Rrightarrow x=\dfrac{2ln3}{ln6}[/tex]

semsee45

Answer:

[tex]\large \text{$ x = \dfrac{2\ln 3}{\ln 6} $}[/tex]

Step-by-step explanation:

[tex]\large \begin{aligned}36^x-7(6^x)-18 & =0\\(6^2)^x-7(6^x)-18 & =0\\(6^x)^2-7(6^x)-18 & =0\\\\\textsf{let }6^x=y\implies y^2-7y-18 & = 0\\(y-9)(y+2) & = 0\\\implies y & = 9, -2\\\\\implies 6^x & = 9, -2\\\textsf{Cannot take logs of -ve numbers} \implies \ln 6^x & = \ln 9 \quad \sf(only)\\x \ln 6 & = 2\ln 3\\x & = \dfrac{2\ln 3}{\ln 6}\\\end{aligned}[/tex]