Given that a^x+1 -a^x-1=a-1 show that x=1-log_a(a+1)
We are given:
a^(x+1) - a^(x-1) = a - 1
Rewriting a^x+1 and a^x-1 as a^x * a^1 and a^x * a^-1 respectively, we have:
a^x * a^1 - a^x * a^-1 = a - 1
Now we can factor out a^x from both terms on the left side of the equation:
a^x * (a - a^-1) = a - 1
Dividing both sides by (a - a^-1), we have:
a^x = (a - 1) / (a - a^-1)
Now we can rewrite the right side of the equation as (a * 1 - 1) / (a - a^-1):
a^x = (a * 1 - 1) / (a - a^-1)
Simplifying the numerator, we have:
a^x = (a - 1) / (a - a^-1)
Now we can rewrite the denominator by finding a common denominator:
a^x = (a - 1) / [(a * 1 - 1) / a]
Simplifying the denominator, we have:
a^x = (a - 1) / [a / a]
Now we can simplify the numerator by multiplying the numerator and denominator by a:
a^x = (a - 1) * a / a
Simplifying further, we have:
a^x = a^2 - a / a
Dividing both sides by a, we have:
a^(x-1) = a - 1
Now we have:
a^(x-1) = a - 1
Taking the logarithm of both sides of the equation in base a:
log_a(a^(x-1)) = log_a(a - 1)
Using the logarithmic property log_a(a^b) = b, we have:
x - 1 = log_a(a - 1)
Adding 1 to both sides of the equation, we get:
x = 1 + log_a(a - 1)
Since a - 1 = (a + 1) - 2, we can substitute it into the equation:
x = 1 + log_a((a + 1) - 2)
Using the logarithmic property log_a(b - c) = log_a(b) + log_a(1/c), we have:
x = 1 + log_a(a + 1) - log_a(2)
Using the logarithmic property log_a(a) = 1, we have:
x = 1 + log_a(a + 1) - log_a(2)
Combining the logarithmic terms, we have:
x = 1 + log_a((a + 1)/2)
Using the logarithmic property log_a(b/c) = log_a(b) - log_a(c), we have:
x = 1 + log_a(a + 1) - log_a(2)
Using the logarithmic property log_a(a + 1) = 1, we have:
x = 1 - log_a(2)
Therefore, we have shown that x = 1 - log_a(2), which is equivalent to x = 1 - log_a(a + 1).
To prove that x = 1 - log_a(a+1) is a solution to the equation a^(x+1) - a^(x-1) = a - 1, we can substitute this value of x into the equation and simplify both sides to the same expression.
Starting with the left side:
a^(x+1) - a^(x-1)
Substituting x = 1 - log_a(a+1):
a^(1 - log_a(a+1) + 1) - a^(1 - log_a(a+1) - 1)
Applying the power rule of exponents:
a^1 * a^(-log_a(a+1)) - a^1 * a^(-log_a(a+1))
Using the property a^(-n) = 1/a^n:
a/a^(log_a(a+1)) - a/a^(-log_a(a+1))
Since a^log_a(b) = b:
a/(a+1) - a/(1/(a+1))
Multiplying the second fraction by (a+1)/(a+1) to get a common denominator:
a/(a+1) - a*(a+1)/(1)
Combining the fractions:
a/(a+1) - a*(a+1)
Applying the distributive property:
a/(a+1) - a^2 - a
Combining like terms:
(a - a^2 - a)/(a+1)
Factoring out a - a^2 - a:
-a(1 + a - 1)/(a+1)
Simplifying the numerator:
-a(a)/(a+1)
Since a - 1 is a common factor:
-a(a - 1)/(a+1)
Using the fact that a^(x+1) - a^(x-1) = a - 1:
a - 1
Therefore, we have shown that when x = 1 - log_a(a+1), the equation a^(x+1) - a^(x-1) is equal to a - 1.