What is x if #ln(3x^2) + ln(x^4) + ln(7) =0#?

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Answer 1 · 2018-03-27T23:06:20

Use the logarithm laws.

#ln(3x^2 * x^4 * 7) = 0#

#21x^6 = e^0#

#x^6 = 1/21#

#x = +-root(6)(1/21)#

Hopefully this helps!

Answer 2 · 2018-03-27T23:14:52

The solutions are #x=+-root6(1/21)#.

(or #x=+-21^(-1/6)#.)

Use this logarithm rule:

#log_color(green)a(color(red)x)+log_color(green)a(color(blue)y)=log_color(green)a(color(red)x*color(blue)y)#

Here's this rule applied to our equation:

#ln(color(red)(3x^2))+ln(color(blue)(x^4))+ln(color(green)7)=0#

#ln(color(red)(3x^2)*color(blue)(x^4))+ln(color(green)7)=0#

#ln(color(red)3color(purple)(x^6))+ln(color(green)7)=0#

#ln(color(red)3color(purple)(x^6)*color(green)7)=0#

#ln(color(brown)21color(purple)(x^6))=0#

#log_e(color(brown)21color(purple)(x^6))=0#

Convert to exponential form:

#e^0=21x^6#

#1=21x^6#

#1/21=x^6#

#root6(1/21)=x#

Since the root is an even power, we add a plus-or-minus sign:

#x=+-root6(1/21)#

#x=+-root6(21^-1)#

#x=+-(21^-1)^(1/6)#

#x=+-21^(-1/6)#

You can check using a graphing calculator:

Since the values of the zeroes are the same as our answer, we are correct. Hope this helped!