How do you solve #7^ { x } - 2\cdot 3^ { x } = 0#?

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Answer 1 · 2017-05-25T11:38:49

#x = frac(ln(2))(ln(7) - ln(3))#

We have: #7^(x) - 2 cdot 3^(x) = 0#

#Rightarrow 2 cdot 3^(x) = 7^(x)#

Let's apply #ln# to both sides of the equation:

#Rightarrow ln(2 cdot 3^(x)) = ln(7^(x))#

Using the laws of logarithms:

#Rightarrow ln(2) + x ln(3) = x ln(7)#

#Rightarrow x ln(7) - x ln(3) = ln(2)#

#Rightarrow x (ln(7) - ln(3)) = ln(2)#

#therefore x = frac(ln(2))(ln(7) - ln(3))#

Answer 2 · 2017-05-25T11:42:43

I got: #x=ln(2)/ln(7/3)=0.81807#

I would first write it as:

#7^x=2*3^x#

and then:

#7^x/3^x=2#

#(7/3)^x=2#

and then apply the natural log on both sides:

#ln(7/3)^x=ln(2)#

#xln(7/3)=ln(2)#

and:

#x=ln(2)/ln(7/3)=0.81807#