How do you evaluate log_a x = 3/2log_a 9 + log_a 2?
3 Answers
We will use the following log properties:
Explanation:
Practice exercises
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Solve for x in the following equation. Round your answers to the nearest hundredth:
log_n2x = 1/3log_n64- 2log3 -
A commonly used logarithm rule is if
a = b -> loga = logb . Solve for x in2^(2x - 3) = 3^(3x)
Good luck!
x = 54
Explanation:
using the
color(blue)" laws of logarithms "
• logx + logy = logxy
• logx - logy = log(x/y)
• logx^n = nlogx hArr nlogx = logx^n
• "If " log_b x = log_b y rArr x = y
rArr log_a x = log_a 9^(3/2) + log_a 2 [ now
9^(3/2) = (sqrt9)^3 = 3^3 = 27 ]
rArr log_a x = log_a (27xx2) = log_a 54 hence x = 54
Explanation:
We are given
Anyways, I now have my equation as
This now gives us
Applying that logic, I take