# Write log6 5 as a logarithm of base 4?

0

Math with no numbers?!?

these are my choices….

a)log4 5/log4 6

b)log4 6/log4 5

c)log5 4/ log6 4

d)log6 4/log5 4

• I’ll use log_b as log base b.

Remember, to change bases use this identity: Log_n(a) = log_m(a)/log_m(n).

Log_6(5) = log_4(5)/log_4(6).

You are right, it is a.

• a million. one hundred twenty five is 5^3, so log5(base) one hundred twenty five is comparable to log5(base) 5^3, and the log5(base) and the 5 cancel out. what’s left is: log5(base) m = (a million/3)3 = a million 5^a million = m m = 5. 2. once you have constants in front of logs, you could deliver them interior logs as exponents. So: a million/4 log sixteen = log (sixteen^a million/4) = log 2 a million/2 log 40 9 = log (40 9^a million/2) = log 7 This makes your equation: log y = log 2 + log 7 = log (2 7) = log (14) y = 14. 3. once you’re including 2 logs with an analogous base, you could multiply their insides (like the two and 7 in the final subject). in this subject: 2 = log6(base) (2 * (b^2 + 2)) = log6(base) (2b^2 + 4) 2b^2 + 4 = 6^2 = 36 2b^2 = 32 b^2 = sixteen b = 4. 4. upload the 2d term to the two factors of the equation to get: log3(base) (5x+5) = log3(base) (x^2 – a million) because of the fact the two factors of the equation have log3(base) in them, they’d in simple terms cancel. 5x + 5 = x^2 – a million x^2 – 5x – 6 = 0 (x – 6)(x+a million) = 0 x = 6 and x = -a million For x = -a million, you get log 0, that’s undefined, so x = -a million isn’t an answer. the only answer left is x = 6.

• I take it the space impies that the number before the space is the log’s base, and the number after it is the number you’re taking the log OF?

Remember that you can change a log into any base you want using this identity:

logbase b = logbase c / logbase c

So yes, ‘a’ is correct.

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