hi !!! LOGARITHM problem ?

Question

if 2^x = 3^y = 6^z , show that z = xy / x + y

Answer 1

Let a = 2^x = 3^y = 6^z

Therefore,

x = log(a) / log(2)
y = log(a) / log(3)
z = log(a) / log(6)

Now, xy = [log(a)]^2 / [log(2) * log(3)]

and x + y = [log(a)][log(2) + log(3)] / [log(2) * log(3)]

therefore, xy / (x + y) = [log(a)]^2 / {[log(a)][log(2) + log(3)]}

Cancelling log(a) and noting that log(2) + log(3) = log(6) gives :

xy / (x + y) = log(a) / log(6), which = z.

Answer 2

2^x = log(2)2^x = x

3^y = log(2)3^y = y log(2)3

6^z = 2^z X 3^z = log(2)2^z + log(2)3^z = z + z log(2)3


thus,
x = y log(2)3
x/y = log(2)3


Subsitude [[ log(2)3 ]] = x/y into x = z + z [[ log(2)3 ]]

x = z + z [[ x/y ]]
x = z ( 1 + x/y )
z = x / ( y/y + x/y)
z = x / [ (y + x) / y ]

z = xy / ( y + x )

writing it down on paper may allow u to see it better

Answer 3

a better approach

2^x = 3^y = 6^z = k

therefore 1/x = log 2/log k; 1/y = log 3/log k and 1/z = log 6/log k

1/x+1/y = (log 2+log 3)/log k = log (6)/log k = 1/z
hence x+y/xy = z hence proved

Answer 4

evaluate the given expression: log base 5 of 5^7 A logarithm is in basic terms an exponent and a base is the quantity it relatively is raised to the flexibility particular in ability of the exponent. So, on your party, we want the logarithm to the backside 5 AND the quantity that we are taking the logarithm of already has 5 as its base (reachable, yet now no longer mandatory) it relatively is raised to the seventh potential, so log5 (5^7) = 7.