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punta
<span>Negro culo
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You can convert (1/625) to an exponent, and it would be ideal to have 5 as the base of it because you want your log base to cancel it out. what i usually do in this case is just test out 5^1, 5^2, etc until i find one that matches the number i need. in this case because the number you're trying to work with is a small fraction, you'll want to use NEGATIVE exponents so it'll create a fraction instead of a large whole number:
5^-1 = 1/5
. . . keep trying those. . .
5^-4 = 1/625
so, because they're equal to one another, it'll be waaay easier after you substitute 5^-4 in place of 1/625
x = log₅ 5⁻⁴
log base 5 of 5 simplifies to 1. subbing in the 5^-4 gets rid of the log for you altogether, and your -4 exponent drops down:
x = -4 is your answer
if the exponent dropping down doesn't make sense to you, you can think of it in another way:
x = log₅ 5⁻⁴
expand the expression so that the exponent moves in front of the log function:
x = (-4) log₅ 5
then, still, log base 5 of 5 simplifies to 1, so you're left with:
x = (-4)1 or x = -4
Hey! If Tanner has t more, just add to to 73, so the expression would be t+73.
<span>Hope that this helped! If it did, please give me a good rating and mark me brainliest for my hard work. Thanks!</span>
