G
Guest
Guest
I am terrible at logs
log5 1/5 =
a. -3
b. -2
c. -1
d. 1
e. 2 is this it?
f. 3
g. 4
h. none of these
log5 1/5 =
a. -3
b. -2
c. -1
d. 1
e. 2 is this it?
f. 3
g. 4
h. none of these
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Hello, adam40g!
Then we have: . \(\displaystyle 5^x \;= \;\frac{1}{5}\)
. . . . . . . . . .or: . \(\displaystyle 5^x \;= \;5^{-1}\)
. . . . . . Hence: . . \(\displaystyle x \;= \;-1\) *
Therefore: . \(\displaystyle \log_5\left(\frac{1}{5}\right) \;= \;-1\) . . . answer (c)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If we have two equal exponential expressions: .\(\displaystyle b^x\ =\ b^y\)
. . and their bases are equal, then the exponents are equal: .\(\displaystyle x\ =\ y\)
Let .\(\displaystyle \log_5\left(\frac{1}{5}\right) \:= \:x\)\(\displaystyle log_5\left(\frac{1}{5}\right)\ =\)
\(\displaystyle a)\;-3\qquad\qquad b)\;-2\qquad\qquad c)\;-1\qquad\qquad d)\;1\qquad\qquad e)\;2\)
\(\displaystyle f)\;3\qquad\qquad g)\;4\qquad\qquad h)\;\text{none of these}\)
Then we have: . \(\displaystyle 5^x \;= \;\frac{1}{5}\)
. . . . . . . . . .or: . \(\displaystyle 5^x \;= \;5^{-1}\)
. . . . . . Hence: . . \(\displaystyle x \;= \;-1\) *
Therefore: . \(\displaystyle \log_5\left(\frac{1}{5}\right) \;= \;-1\) . . . answer (c)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If we have two equal exponential expressions: .\(\displaystyle b^x\ =\ b^y\)
. . and their bases are equal, then the exponents are equal: .\(\displaystyle x\ =\ y\)