How do you solve #4^x - 2^x = 0#?

Question

Theodore Amidon · Accepted Answer

#x=0# #4^x-2^x=0# now #4=2^2=>4^x=2^(2x)# #4^x-2^x=0=>2^(2x)-2^x=0# Factorization #2^x(2^x-1)=0# either # 2^x=0=>#no real solns or #2^x-1=0# #2^x=1=>x=0#

Sadie Ackerman · Answer

#x=0#

#4^x-2^x=0#

#4^x=2^x#

On both sides, apply the natural logarithm:

#ln(4^x)=ln(2^x)#

Recalling that #ln(a^b)=bln(a)#:

#xln(4)=xln(2)#

So, #x=0# is obviously a solution as it will result in #0=0.#

Beyond that, there are no solutions as #x# cancels out and we're left with

#ln(4)=ln(2)# which is not true.

Serenity Johnson · Answer

The answer is #x=0#.

The other solutions on the page are accurate; I just wanted to show you another way to tackle this issue: #4^x-2^x=0# #4^x=2^x# #(2^2)^x=2^x# #2^(color(blue)(2x))=2^color(blue)x# Since the bases are equal, it follows that the exponents should be as well: #color(blue)(2x)=color(blue)x# #x=0#

Benjamin Achtenberg · Answer

Real solution: #x=0#

Complex solutions: #x = (2kpii)/ln 2" "# for any integer #k#.

Given: #4^x-2^x = 0# Note that #4^x = (2 * 2)^x = 2^x * 2^x# Thus, we have: #0 = 4^x-2^x = 2^x(2^x-1)# Note that #2^x = 0# has no solutions (real or complex). So: #2^x = 1# This has real solution #x = 0# Consider intricate fixes. Note that #e^(2pii) = 1#, so #e^(2kpii) = 1# for any integer #k#. These are the only complex values for which #e^t = 1#. Thus, we discover: #e^(2kpii) = 1 = 2^x = (e^(ln 2))^x = e^(x ln 2)# So: #x ln 2 = 2kpii# So: #x = (2kpii)/ln 2" "# for any integer #k#

Theodore Amidon · Answer

undefined To solve the equation $4^x - 2^x = 0$, we can factor out $2^x$ to get:

$$2^x(2^x - 1) = 0$$

Now, using the zero-product property, we set each factor equal to zero:

1. $2^x = 0$

This equation has no real solutions because $2^x$ is always positive.

2. $2^x - 1 = 0$

Solving this equation:

$$2^x = 1$$

Since $2^0 = 1$, the solution to this equation is $x = 0$.

Therefore, the only solution to the equation $4^x - 2^x = 0$ is $x = 0$.