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

[Yankel]

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 .

 

[HOI]

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.

 

[Yankel]

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.

 

[Greg]

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

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

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

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

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

$$18^3=18^{2x}\implies x=\frac32$$