By letting x=5^y, show that the equation p5^y + 5^-y = 5 can also be written as px^2-5x+1 = 0
p5^y + 5^-y = 5
p5^y + 1/5^y = 5
now replace
px + 1/x = 5
times x
px^2 + 1 = 5x
px^2 - 5x + 1 = 0
which is what we want
To show that the equation p(5^y) + (5^(-y)) = 5 can be written as px^2 - 5x + 1 = 0, we will use the substitution x = 5^y.
1. Start with the given equation: p(5^y) + (5^(-y)) = 5.
2. Substitute x = 5^y: p(x) + 1/x = 5.
3. Multiply through by x to clear the denominator: px + 1 = 5x.
4. Subtract 5x from both sides: px - 5x + 1 = 0.
5. Rearrange the terms: px^2 - 5x + 1 = 0.
We have now shown that the equation p(5^y) + (5^(-y)) = 5 can indeed be written as px^2 - 5x + 1 = 0 by substituting x = 5^y.
To show that the equation p5^y + 5^-y = 5 can also be written as px^2 - 5x + 1 = 0, we can substitute x = 5^y into the equation and simplify.
Let's start by substituting x = 5^y into the equation p5^y + 5^-y = 5:
p(5^y) + (5^(-y)) = 5
Since x = 5^y, we can replace 5^y with x in the equation:
px + (5^(-y)) = 5
Now, we need to get rid of the term with 5^(-y) by expressing it in terms of x. Notice that 5^(-y) is the same as (1/5)^y. Since we have x = 5^y, we can write (1/5)^y as (1/5)^x:
px + (1/5)^x = 5
Now, let's simplify the equation further. We can multiply each term by 5^x to eliminate the fractions:
px(5^x) + (5^x)(1/5)^x = 5(5^x)
This simplifies to:
px(5^x) + 1 = 5(5^x)
Next, we can rewrite 5 as (5^1) to get:
px(5^x) + 1 = (5^1)(5^x)
Using the rule of exponents that states when multiplying two terms with the same base, you add their exponents, we can simplify the right side of the equation:
px(5^x) + 1 = 5^(1+x)
Finally, we can write 5^(1+x) as (5^1)(5^x) using the same exponent rule:
px(5^x) + 1 = (5^1)(5^x)
Simplifying further, we have:
px(5^x) + 1 = 5^x+1
Now, let's rearrange the terms to get the equation in the desired form:
px(5^x) - 5^x + 1 = 0
Since x = 5^y, we can substitute x back into the equation, giving us:
p(5^y)(5^y) - (5^y) + 1 = 0
This simplifies to:
p(5^2y) - 5^y + 1 = 0
And finally, using the property that (5^2y) is the same as (5^y)^2, we get:
px^2 - 5x + 1 = 0
So, we have shown that the equation p5^y + 5^-y = 5 can also be written as px^2 - 5x + 1 = 0 by substituting x = 5^y and simplifying the expression.