One way to prove 1/x + 1/z = 2/y with the given information is to use logarithms. By taking the logarithm in base 10, if f = k^r, then

log f = log k^r. The power (r) is moved to the front of the log, so:

log f = r...

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One way to prove 1/x + 1/z = 2/y with the given information is to use logarithms. By taking the logarithm in base 10, if f = k^r, then

log f = log k^r. The power (r) is moved to the front of the log, so:

log f = r log k.

Using the previous thought about logarithms with the given information.

a^x = b^y = c^z

log a^x = log b^y = log c^z

**x log a = y log b = z log c** (Equation 1)

Starting with 1/x + 1/z, since x = (z log c)/ log a, then

1/x = log a/(z log c) and 1/z = log c/(x log a)

1/x + 1/z = log a/(z log c) + log c/(x log a)

Using equation 1, x log a = z log c. The denominators are equal.

= log a/(z log c) + log c/(z log c)

= `(log a + log c)/(z log c)`

`=(log a*c)/(zlogc)` Using the rule that log a*c = log a + log c

`=(log b^2)/(y log b)` a*c = b^2 and z log c = y log b

`=(2logb)/(ylogb)=2/y`

The proof uses the following logarithm rules

log x^y = y log x

log x*y = log x + log y