SOLUTION: Using log rules
Loga = 0.3
Logb = -0.5
Solve
log(1000a^2)
log(4throot √(b/a))
I need to evaluate while showing the use of log rules
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Solve
log(1000a^2)
log(4throot √(b/a))
I need to evaluate while showing the use of log rules
=======================================
What I have is: log(1000)+log(a^2)
wouldn't log(1000) be the same as log[Base 10] 10 so would that count as the evaluation for that part?
then log(0.3^2) = log(0.09)
therefore, the answer log[base 10]10 + log(0.09)
and log(4th root √(b/a)) would be log(b/a)^(1/4) and therefore log(-0.5)^(1/4) - log(0.3)^(1/4)
I feeel very unsure about these answers please help
You can put this solution on YOUR website! Using log rules
Loga = 0.3
Logb = -0.5
Solve
log(1000a^2)
= log(1000) + log(a^2)
= log(1000) + 2*log(a)
Base 10 is the default, and 1000 = 10^3
----
= 3 + 2*0.3
= 3.6
===============================
log(4throot √(b/a))
= (1/4)*log(b/a)
= (1/4)
= (1/4) = (1/4)*(-0.8)
= -0.2
====================================
I need to evaluate while showing the use of log rules
=======================================
What I have is: log(1000)+log(a^2)
wouldn't log(1000) be the same as log[Base 10] 10 so would that count as the evaluation for that part?
then log(0.3^2) = log(0.09)
therefore, the answer log[base 10]10 + log(0.09)
and log(4th root √(b/a)) would be log(b/a)^(1/4) and therefore log(-0.5)^(1/4) - log(0.3)^(1/4)
I feeel very unsure about these answers please help
