Solve 8^x+8^(-x)=? When 4^x+4^(-x)=8

  • MHB
  • Thread starter Monoxdifly
  • Start date
In summary, using the substitution method, we can find that 8^x+8^{-x} is equal to 7 times the square root of 10. However, this answer is not listed among the options provided.
  • #1
Monoxdifly
MHB
284
0
If \(\displaystyle 4^x+4^{-x}=8\), then \(\displaystyle 8^x+8^{-x}=?\)
A. 14
B. 15
C. 16
D. 17
E. 18

What should I do? I tried substituting \(\displaystyle y = 4^x\) but it didn't help since the quadratic equation formed didn't have real roots.
 
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  • #2
Monoxdifly said:
If \(\displaystyle 4^x+4^{-x}=8\), then \(\displaystyle 8^x+8^{-x}=?\)
A. 14
B. 15
C. 16
D. 17
E. 18

What should I do? I tried substituting \(\displaystyle y = 4^x\) but it didn't help since the quadratic equation formed didn't have real roots.

we know $4=2^2$ and $8= 2^3$ so put $2^x = y$
we get $y^2 + \frac{1}{y^2} = 8\cdots(1)$
we need to find $y^3+ \frac{1}{y^3}$
from (1) we get
$(y + \frac{1}{y})^2 -2 = 8$
or $(y + \frac{1}{y}) = \sqrt{10}$

now you should be able to proceed.
 
  • #3
kaliprasad said:
we know $4=2^2$ and $8= 2^3$ so put $2^x = y$
we get $y^2 + \frac{1}{y^2} = 8\cdots(1)$
we need to find $y^3+ \frac{1}{y^3}$
from (1) we get
$(y + \frac{1}{y})^2 -2 = 8$
or $(y + \frac{1}{y}) = \sqrt{10}$

now you should be able to proceed.

Well...
\(\displaystyle (y + \frac{1}{y})^3 = (\sqrt{10})^3\)
\(\displaystyle y^3+3y^2(\frac{1}{y})+3y(\frac{1}{y^2})+\frac{1}{y^3}=10\sqrt{10}\)
\(\displaystyle y^3+\frac{1}{y^3}+3y+3(\frac1y)=10\sqrt{10}\)
\(\displaystyle y^3+\frac{1}{y^3}+3(y+\frac1y)=10\sqrt{10}\)
\(\displaystyle y^3+\frac{1}{y^3}+3\sqrt{10}=10\sqrt{10}\)
\(\displaystyle y^3+\frac{1}{y^3}=7\sqrt{10}\)
Not in the options...
 
  • #4
Monoxdifly said:
Well...
\(\displaystyle (y + \frac{1}{y})^3 = (\sqrt{10})^3\)
\(\displaystyle y^3+3y^2(\frac{1}{y})+3y(\frac{1}{y^2})+\frac{1}{y^3}=10\sqrt{10}\)
\(\displaystyle y^3+\frac{1}{y^3}+3y+3(\frac1y)=10\sqrt{10}\)
\(\displaystyle y^3+\frac{1}{y^3}+3(y+\frac1y)=10\sqrt{10}\)
\(\displaystyle y^3+\frac{1}{y^3}+3\sqrt{10}=10\sqrt{10}\)
\(\displaystyle y^3+\frac{1}{y^3}=7\sqrt{10}\)
Not in the options...

Wolfram confirms that your solution is correct.
So it appears there is either a typo in the problem statement, or the correct answer is indeed not listed.
 
  • #5
Ah, okay then. Thanks to both of you.
 

FAQ: Solve 8^x+8^(-x)=? When 4^x+4^(-x)=8

1. What is the value of x in the equation 8^x+8^(-x)=? When 4^x+4^(-x)=8

The value of x in this equation cannot be determined without further information. Both equations have the same solution, but there are multiple values of x that satisfy the equation.

2. Is there a specific method for solving this type of equation?

Yes, there are several methods for solving equations with variables in the exponent. Some common methods include taking the logarithm of both sides, using substitution, or graphing the equations to find the intersection point.

3. Can this equation be solved algebraically?

Yes, this equation can be solved algebraically using the methods mentioned in the previous answer. However, the solution may involve complex numbers.

4. How many solutions does this equation have?

This equation has an infinite number of solutions. Since there are two variables (x and y) and only one equation, there are multiple combinations of x and y that can satisfy the equation.

5. Can this equation be solved without using a calculator?

Yes, this equation can be solved without a calculator using algebraic methods. However, depending on the values of x and y, the solutions may involve irrational or complex numbers that may be difficult to calculate without a calculator.

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