Exponential Equation solve 54⋅2^(2x)=72^x⋅√0.5

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SUMMARY

The exponential equation \(54\cdot 2^{2x}=72^{x}\cdot \sqrt{0.5}\) can be solved by manipulating the bases and exponents. The equation simplifies to \(54\cdot 4^x=18^{x}\cdot4^x\cdot\sqrt{\frac{1}{2}}\), leading to the conclusion that \(x=\frac{3}{2}\). The key steps involve recognizing the relationships between the numbers and ensuring the exponents are equal, despite the initial confusion regarding the exponents. The use of logarithms is explicitly prohibited in this solution.

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Yankel
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Hello all,

I need assistance in solving this exponential equation.

\[54\cdot 2^{2x}=72^{x}\cdot \sqrt{0.5}\]

The final answer should be 3/2.

My strategy was to try and bring to a state where the exponents are equal. We know that 54 is 6 times 9. We also know 72 is 8 times 9. The solution probably involves the fact that 9 appears in both numbers.

Can you kindly assist ? Oh, one more thing, important thing. The use of logarithms is forbidden... :-)

Thank you .
 
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Yankel said:
Hello all,

I need assistance in solving this exponential equation.

\[54\cdot 2^{2x}=72^{x}\cdot \sqrt{0.5}\]

The final answer should be 3/2.

My strategy was to try and bring to a state where the exponents are equal. We know that 54 is 6 times 9.
More to the point, 54 is 2 times 27: 54= 2(3^3)

We also know 72 is 8 times 9.
Yes, and that is 72= (2^3)(3^2). Of course \sqrt{0.5}= \frac{1}{\sqrt{2}}= 2^{-1/2}

The solution probably involves the fact that 9 appears in both numbers.

Can you kindly assist ? Oh, one more thing, important thing. The use of logarithms is forbidden... :-)

Thank you .
54(2^{2x})= 2(3^3)(2^{2x})= 2^{2x+1}(3^3) and 72^x\sqrt{0.5}= 2^x(3^{2x})2^{-1/2}= 2^{x- 1/2}3^{2x}

54(2^{2x})=72^x\sqrt{0.5} is the same as
2^{2x+1}(3^3)= 2^{x- 1/2}3^{2x}

But now we have a problem! In order for those to be equal the exponents of both 2 and 3 must be the same on each side. We must ave both 2x+ 1= x- 1/2 and 3= 2x. To solve 2x+ 1= x- 1/2, subtract x and 1 from both sides: x= -3/2. To solve 3= 2x divide both sides by 2: x= 3/2. Those are NOT the same! There is no value of x that satisfies this.
 
Thank you very much.

I think that you have a small mistake with the exponents at the beginning but the general approach helped me get to the correct solution.
 
$$54\cdot 2^{2x}=72^{x}\cdot \sqrt{0.5}$$

$$54\cdot 4^x=18^{x}\cdot4^x\cdot\sqrt{\frac12}$$

$$54=18^x\cdot\sqrt{\frac12}$$

$$2\cdot54^2=18^{2x}$$

$$2\cdot3^2\cdot18^2=18^{2x}$$

$$18^3=18^{2x}\implies x=\frac32$$
 

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