If #216 * 3^(2x)-73*3^x-3=0# and #u=3^x#, what is the value of #u#?

2 Answers
Oct 5, 2017

#u=0.375#

Explanation:

If #color(blue)u=color(magenta)(3^x)#
then
#color(white)("xxx")216 * color(magenta)(3^(color(black)2x))+(-73) * color(magenta)(3^x)-3=0#
becomes
#color(white)("xxx")216color(blue)u^2-73color(blue)u-3=0#

Using the quadratic formula (and a calculator)
#color(white)("xxx")color(blue)u=(73+-sqrt((-73)^2-4 * 216 * (-3)))/(2 * 216)#

#color(white)("xxx")=0.375#
or
#color(white)("xxx")=-0.037bar(037)#

...but if #u=3^x# then #u > 0, AA x in RR#
and therefore the negative version is extraneous,
leaving only
#color(white)("xxx")color(blue)u=0.375#

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As a bonus we could find the value of #x# (approximately)

If #3^x=u#
then
#color(white)("xxx")log_3(3^x)=log_3(u)#

#color(white)("xxx")x=log_3(u)#

#color(white)("xxx")x=log_3(0.375)#

then using a calculator
#color(white)("xxx")x~~-0.892789261#

Oct 5, 2017

#x= -0.89#

Explanation:

The equation is in the form, #au^2-bu-c# where #u=3^x#.
Since it is difficult to use #x# straightaway to solve the equation, we have replaced it with ā€˜uā€™.
Once we solve for u, we will then replace u with 3^x and find the value of x.

#216u^2-73u-3=0#
#216*3=648# can be factorised as #81*8#
#216u^2+8u-81u-3=0#
#8u*(27u+1)-3*(27u+1)=0#
#(27u+1)(8u-3)=0#
#u=-(1/27) & (3/8)#

#3^x=-(1/27)=-3^-3# This solution is imaginary.

#3^x=3/8#
#xlog3=log(3/8)=log(3)-log(8)#
#x=(log(3)/log(3))-(log(8)/log(3))=1-1.89=-0.89#