How do you expand #log(x(x^3+9)^(-1/2))#?

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Answer 1 · 2016-03-20T16:06:04

We must use the log properties to simplify.

1) Use #log_n(a xx b) = log_na + log_nb# to seperate the expression into two logarithmic expressions.

= #logx + log(x^3 + 9)^(-1/2)#

2). Use #loga^x = xloga# to get rid of the exponent.

Since the exponent is negative, we should first use the exponential property #a^(-n) = 1/(a^n)#

=#logx + log(1/(x^3 + 9)^(1/2))#

=#logx + 1/2log(1/(x^3 + 9))#

3). We now use the rule #log_a(n/m) = log_an - log_am#

=#logx + 1/2(log1 - log(x^3 + 9))#

=#logx + 1/2log1 - 1/2log(x^3 + 9)#

Since #log1 = 0#, we are left with #logx - 1/2log(x^3 + 9)#

We could have just used the negative #-1/2# at step 2, but I took the opportunity to show you an additional log rule.

Hopefully this helps!