Verified answer

Answer:

[tex]\qquad\qquad\boxed{ \sf{ \:\bf \: x \: \approx \: 1.187 \: }} \\ \\ [/tex]

Step-by-step explanation:

Given equation is

[tex]\sf \: {4}^{x} + {6}^{x} = {9}^{x} \\ \\ [/tex]

On dividing each term by [tex]6^x [/tex], we get

[tex]\sf \: \dfrac{ {4}^{x} }{ {6}^{x} } + 1 = \dfrac{ {9}^{x} }{ {6}^{x} } \\ \\ [/tex]

[tex]\sf \: {\bigg(\dfrac{2}{3} \bigg) }^{x} + 1 = {\bigg(\dfrac{3}{2} \bigg) }^{x} \\ \\ [/tex]

[tex]\sf \: Let \: assume \: that \: {\bigg(\dfrac{3}{2} \bigg) }^{x} = y \\ \\ [/tex]

So, above equation can be rewritten as

[tex]\sf \: \dfrac{1}{y} + 1 = y \\ \\ [/tex]

[tex]\sf \: \dfrac{1 + y}{y} = y \\ \\ [/tex]

[tex]\sf \: {y}^{2} = y + 1 \\ \\ [/tex]

[tex]\sf \: {y}^{2} - y - 1 = 0 \\ \\ [/tex]

So, using quadratic formula, we get

[tex]\sf \: y = \dfrac{ - ( - 1) \: \pm \: \sqrt{ {( - 1)}^{2} - 4( - 1)(1) } }{2(1)} \\ \\ [/tex]

[tex]\sf \: y = \dfrac{ 1 \: \pm \: \sqrt{1 + 4 } }{2} \\ \\ [/tex]

[tex]\sf \: y = \dfrac{ 1 \: \pm \: \sqrt{5} }{2} \\ \\ [/tex]

[tex]\sf \: y = \dfrac{ 1 \: + \: \sqrt{5} }{2} \: \: or \: \: y = \dfrac{ 1 \: - \: \sqrt{5} }{2} \\ \\ [/tex]

[tex]\sf \: {\bigg(\dfrac{3}{2} \bigg) }^{x}= \dfrac{ 1 \: + \: \sqrt{5} }{2} \: \: or \: \: {\bigg(\dfrac{3}{2} \bigg) }^{x} = \dfrac{ 1 \: - \: \sqrt{5} }{2} \\ \\ [/tex]

[tex]\sf \: {\bigg(\dfrac{3}{2} \bigg) }^{x}= \dfrac{ 1 \: + \: 2.236 }{2} \: \: or \: \: {\bigg(\dfrac{3}{2} \bigg) }^{x} = \dfrac{ 1 \: - \:2.236 }{2} \\ \\ [/tex]

[tex]\sf \: {\bigg(\dfrac{3}{2} \bigg) }^{x}= \dfrac{ 3.236 }{2} \: \: or \: \: {\bigg(\dfrac{3}{2} \bigg) }^{x} = \dfrac{ - \:1.236 }{2} \\ \\ [/tex]

[tex]\sf \: {\bigg(\dfrac{3}{2} \bigg) }^{x}= \dfrac{ 3.236 }{2} \: \: or \: \: {\bigg(\dfrac{3}{2} \bigg) }^{x} = \dfrac{ - \:1.236 }{2} \: \: \: \{rejected \} \\ \\ [/tex]

[tex]\sf \: {\bigg(\dfrac{3}{2} \bigg) }^{x}= 1.618 \: \\ \\ [/tex]

On taking log on both sides, we get

[tex]\sf \:log {\bigg(\dfrac{3}{2} \bigg) }^{x}= log(1.618 )\: \\ \\ [/tex]

[tex]\sf \:x \: log {\bigg(\dfrac{3}{2} \bigg) }^{}= log(1.618 )\: \\ \\ [/tex]

[tex]\sf \:x \: log {\bigg(1.5 \bigg) }^{}= 0.209\: \\ \\ [/tex]

[tex]\sf \: \: 0.1761x \: = \: 0.209\: \\ \\ [/tex]

[tex]\sf \: x = \dfrac{0.209}{0.1761} \\ \\ [/tex]

[tex]\sf \: x =1.1868 \\ \\ [/tex]

[tex]\bf\implies \: x \: \approx \: 1.187 \\ \\ [/tex]

Divide both sides by 4x:

1+(32)x=(94)x

Note that

94=(32)2

Put y=(32)x:

1+y=y2

And now solve for y:

y=1±5–√2

By definition, y>0, therefore

y=1+5–√2

and

(32)x=1+5–√2

Apply the logarithm

xln(32)=ln(1+5√2)

and finally

x=ln(1+5√2)ln(32)≈1.1868