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Algebra help, Logarithms. Help!?
I just don't understand these! And I have a quiz tomorrow that I must pass:( If you could work these and explain how to do them, that would be great!
Use log(down5) 2=0.4307 and log(down5) 3=0.6826 to approximate the value of each expression.
1.) log(down5) 18
2.) log(down5) 5/2
For these questions, solve each equation or inequality. If necessary, round to four decimal places.
1.) log(down5) n=1/3 log(down5) 64 + 1/2 log(down5) 49
2.) log(down6) (5-2a) - log(down6) (3a) = 1
3.) log(down3) (x-3) + log(down3) (x+2) = log(down3) 6
If you can answer some of these, that would be awesome! I really need help understanding them :-(
2 Answers
- 1 decade agoFavorite Answer
its called base log_5(2) would be a nicer way to put this.
Know your properties of logs 1) log_5(18) =3 log(3)=3*0.6826
log multiplication outside is addition inside, log division is subtraction on inside, log(x)+log(y) = log(x*y); log(x/y) = log(x)-log(y), as long as they have the same base, log(x)+log(x) = 2*log(x)= log(x^2)
- 1 decade ago
You're basically finding to what exponent the number after log equals the number you put in down. For instance, log(down5) 25 would equal 2, because five squared equals 25. However, I don't know how to solve them without whole-numbered exponents.
