if a^x=(a/k)^y=k^m and a is not equal to 1 then show that 1/x-1/y=1/m
show me work plz
(a/k)^y=k^m
take log of both sides:
y(log (a/k) = mlogk
y(log a - logk = mlogk
yloga = mlogk + ylogk
yloga = logk (m+y)
logk = yloga/(m+y)
also:
a^x = k^m
xloga = mlogk
logk = xloga/m
xloga /m = yloga/(m+y)
divide by loga
x/m = y/(m+y)
ym = x(m+y)
ym = xm + xy
ym - xm = xy
m(y-x) = xy
m = xy/(y-x)
m = x(x+y)/y
we are supposed to show that 1/x - 1/y = 1/m
1/m = (y-x)/(xy)
m = xy/(y-x), which is what I have done above.
checking:
let y = 6, x = 4, then m = 12
and if a^x = k^m
a^4 = k^12
4loga = 12logk
log a = 3logk
log a = log k^3
a = k^3, let k = 2, then a = 8
checking the question:
a^x=(a/k)^y=k^m
8^4 = 4096
(a/k)^y =(8/2)^6 = 4^6 = 4096
k^m = 2^12 = 4096
but
1/4 - 1/6 = 1/12, illustrating the result
Thankyou somuch
Thank you so much
To solve this problem, we'll start by taking the logarithm of both sides of the equation. Since the base of the logarithm can be any positive value, we'll choose the common logarithm (log base 10) for simplicity.
Step 1: Take the logarithm of both sides:
log(a^x) = log((a/k)^y) = log(k^m)
Using the property of logarithms, which states that log(a^b) = b * log(a), we can rewrite the equation:
x * log(a) = y * log(a / k) = m * log(k)
Step 2: Rearrange the equation to solve for x:
x = (y * log(a / k)) / log(a)
Step 3: Rearrange the equation to solve for y:
y = (x * log(a)) / log(a / k)
Step 4: Rearrange the equation to solve for m:
m = (x * log(a)) / log(k)
Now, let's substitute these values back into the equation 1/x - 1/y = 1/m and simplify:
1/x - 1/y = 1/m
Substituting the expressions for x, y, and m:
1 / ((y * log(a / k)) / log(a)) - 1 / ((x * log(a)) / log(a / k)) = 1 / ((x * log(a)) / log(k))
Next, let's simplify the fractions by inverting and multiplying:
log(a) / (y * log(a / k)) - log(a / k) / (x * log(a)) = log(k) / (x * log(a))
Now, let's find a common denominator:
(log(a) * log(a / k) - (log(a / k))^2) / (x * y * log(a / k)) = log(k) / (x * log(a))
To proceed further, let's focus on the left side of the equation. We can simplify the numerator and denominator:
(log(a) * log(a / k) - (log(a / k))^2) / (x * y * log(a / k))
Expanding (log(a / k))^2 gives:
(log(a) * log(a / k) - log(a / k) * log(a / k)) / (x * y * log(a / k))
Notice that (log(a / k))^2 can be factored out:
(log(a / k))(log(a) - log(a / k)) / (x * y * log(a / k))
Canceling out the common factor of (log(a / k)), we are left with:
(log(a) - log(a / k)) / (x * y)
Using the logarithmic property log(a) - log(b) = log(a / b), we get:
log(a / (a / k)) / (x * y)
Simplifying the numerator gives:
log(k) / (x * y)
The left side of the equation now becomes:
log(k) / (x * y) = log(k) / (x * log(a))
Now, notice that log(k) appears on both sides of the equation. We can cancel it out:
1 / (x * y) = 1 / (x * log(a))
Finally, we can rewrite the equation as:
1 / x - 1 / y = 1 / m
Therefore, we have shown that 1/x - 1/y = 1/m.