Apply algebraic reasoning to show that a=b^(loga/logb) for any a,b>0
take the log of each side
loga=(loga/logb)logb
now reduce.
I have:
loga=(loga/logb)logb
(loga/logb )(1/loga)=logb
(1/logb)=logb
Ok now what?
you erred.
loga=(loga/logb)logb
the logb on the right side divide out (one on numerator, one in denominator)
loga=loga
divide both sides by loga.
To prove that a = b^(loga/logb) for any a, b > 0, we will use algebraic reasoning. Here's how to approach this:
Step 1: Start with the given equation: a = b^(loga/logb)
Step 2: Take the logarithm of both sides of the equation. You can use any base for the logarithm, but it's convenient to use the same base as the logarithm in the exponent. In this case, we'll use base b logarithm.
logb(a) = logb(b^(loga/logb))
Step 3: Apply the logarithmic property: logb(x^y) = y * logb(x)
logb(a) = (loga/logb) * logb(b)
Step 4: Simplify the right side of the equation. logb(b) is equal to 1.
logb(a) = loga/logb
Step 5: Multiply both sides of the equation by logb to isolate logb(a).
logb(a) * logb = loga
Step 6: Apply the exponential property. Since logb(b) = 1, we have b^logb(a) = a.
b^(logb(a) * logb) = a
Step 7: Use the commutative property of multiplication to rearrange the expression.
b^(loga/logb) = a
Step 8: Therefore, we have proved that a = b^(loga/logb) for any a, b > 0 using algebraic reasoning.
Please note that throughout this explanation, we have assumed all logarithms to be positive, which is true when a and b are positive and greater than 1.