How do you solve #log_5 2x − 5 = −4#?

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Answer 1 · 2016-05-04T17:39:11

#x=1563/625#

Use definition #log_b x=y iff b^y=x#

#5^-4 = 2x-5#

#1/5^4 = 2x-5#

#1/625 = 2x-5#

#1/625 + 5 = 2x#

#3126/625 = 2x#

#x=1563/625#