How do you solve #3^(2x) – 2*3^(x+5) + 3^10 = 0#?

2 Answers
May 18, 2016

Answer:

#x=5#

Explanation:

Remember the polynomial identity
#(a+b)^2=a^2+2ax+b^2#. Choosing #a=3^x# and #b=-3^5# we have
#(3^x-3^5)^2=3^{2x}-2*3^x*3^5+3^{10}#. So, our problem is reduced to: Solve for #x# the condition #(3^x-3^5)^2=0->3^x-3^5=0->3^x=3^5->x=5#

May 18, 2016

Answer:

x=5

Explanation:

At first sight this seems a quadratic equation of the form:
#y^2+by+c# where #y=3^x#. So let's transform the equation into the quadratic form:

#3^(2x)-2xx3^5xx3^x+3^10#

We can see that the equation can also be reduced to

#(3^x)^2-2xx3^5xx3^x+(3^5)^2#

Remember that :

#color(blue)(a^2-2ab+b^2=(a-b)^2)#

The equation can be arranged into:

#(3^x-3^5)^2=0#

So:

#3^x-3^5=0#

#3^x=3^5#

#x=5#