How do I solve #log_sqrt5##0.2#?

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Answer 1 · 2017-10-14T07:35:56

# -2.#

We know that, #log_a y=x iff a^x=y.#

#"So, if, "log_sqrt5 0.2=x," then, "(sqrt5)^x=0.2.#

#:. (5^(1/2))^x=0.2=2/10=1/5.#

#:. 5^(x/2)=5^(-1).#

#:. x/2=-1.#

#:. x=-2.#

# rArr log_sqrt5 0.2=-2.#