Question

simplify
5^5log1/5x

Answer 1

to better write it:
\[5^{5\log_{1/5}x}\]

Answer 2

using.. log property:
\[a ^{x} = y\]
\[\log_{a} y = x\]

I cannot solve any further, so need HELP! Can someone please do :(

Answer 3

....

Answer 4

what?????

Answer 5

first,we need to get the base of the logarithm to be the same as the base of the exponentiation.

Answer 6

Let y = log(base 1/5) x

Answer 7

Then

(1/5)^y = (1/5)^[log(base 1/5) x] = x

Answer 8

Take log base 5 of both sides:

log5[(1/5)^y] = log5(x)

Answer 9

Using the log law log(a^b) = b log(a),

y log5(1/5) = log5(x)

Answer 10

But 1/5 = 5^(-1), so log5(1/5) = -1

Put this into the equation:

-y = log5(x)
y = -log5(x) = log5(x^(-1)) = log5(1/x)

Answer 11

5^(5log1/5(x))
say y = 5log1/5(x)
then y = 5log(x)/log(1/5) ( some fictious base)
y=-5log(x)/log(5)
y=-5log5(x) (i.e. log(x) with base 5)
y=log5(x^-5)
then 5^y=5^(log5(x^-5) = x^-5=1/x^5
answer is 1/x^5

Answer 12

So log(1/5)(x) = log5(1/x)
5 log(1/5)(x) = 5 log5(1/x) = log5[(1/x)^5]
5^[5 log(1/5)(x)] = 5^{ log5[(1/x)^5] }
= (1/x)^5 = x^(-5)

Answer 13

satisfied?

Answer 14

(1/5)^y = (1/5)^[log(base 1/5) x] is equal to x?

Answer 15

the second log property you wrote is being applied there

Answer 16

no, sorry, i am MORE confused! Is there a simpler way? or can you use the 'Equation'?

Answer 17

the main formula used is
logb(a) = log(a)/log(b). then you can easily understand the rest.

Answer 18

but 1/5 is the base of the log in the question..
how did you use the above mentioned property.. this property is not for log bases but log values!