write the equation in it's equivalent logarithmic form. 5^-2=1/25?

c. log to the base 5(1/25)=-2 answer @(^_^)@

log (base 4) of sixty 4 = 3 How did you comprehend what is going the position? First write the equation in exponential style: 4³ = sixty 4 in this equation, there are 2 things you may comprehend, the bottom and the exponent. (the 0.33 element is the final merchandise contained in the equation, i.e. sixty 4) In logarithmic style: its continuously log with a particular base of a few variety, the position you'll discover the bottom in subscript after the log, and then the range you elect to discover the log of. What you elect to entice close in log style is that the log evaluates to an exponent. E.g.: log (base 3) 80 one = 4 the log evaluated to 4, that is the exponent of base 3 to get 80 one. So, to sum it up: In exponential style: Base^exponent = the different variety In log style: log (with Base) of the different variety = exponent desire the reply wasn't too perplexing

b^x = y → log_b(y) = x

5^-2 = 1/25
log_5(1/25) = -2
(answer c)

Lx(z) = y => x^y = z
Lb(-z) = c => x^y = -z
a.
L5 (-2) = L5(1/25)
L5 (-2) = L5(1) - L5(25)

b.
L1/5 (5) = L1/5(-2 )


u can refine it
c.
L5(1/25) = L5(-2)
L5(1) - L5(25) = L5(-2)

d.
L-2(1/25) = L-2(5)
L-2(1) - L-2(25) = L-2(5)

C is your intended correct answer; it should be written thus:

c. log to the base 5 of [ 1/25 ] = - 2
- - - - - - - -
In general B^x = N would be the exponential form of this logararithmic form:

log, base B, of N is x

its letter C.

Recall

Log_b a = c
which may be written to
b^c = a


so
5^-2=1/25
Log_5 (1/25) = -2

c. use pigtails