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

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In summary, I was able to solve this equation by using the exponents and the fact that 9 appears twice.
  • #1
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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  • #2
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: [tex]54= 2(3^3)[/tex]

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

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

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

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.
 
  • #3
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.
 
  • #4
$$54\cdot 2^{2x}=72^{x}\cdot \sqrt{0.5}$$

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

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

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

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

$$18^3=18^{2x}\implies x=\frac32$$
 

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

1. How do you solve an exponential equation?

To solve an exponential equation, you need to isolate the variable in the exponent on one side of the equation by using logarithms.

2. What is the first step in solving an exponential equation?

The first step in solving an exponential equation is to make sure that the bases on both sides of the equation are the same.

3. How do you solve an equation with different bases?

You can solve an equation with different bases by using the property of logarithms which states that logb(xy) = y*logb(x). This allows you to rewrite the equation with the same base on both sides.

4. What is the purpose of using logarithms in solving exponential equations?

The purpose of using logarithms in solving exponential equations is to convert the equation into a linear form, where the variable is not in the exponent. This makes it easier to solve for the variable.

5. How do you solve an exponential equation with a radical?

To solve an exponential equation with a radical, you need to first isolate the radical on one side of the equation. Then, you can raise both sides of the equation to the reciprocal power of the radical to eliminate it. From there, you can use logarithms to solve for the variable.

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