How do you write an equation of a line going through #2^x-5

Question

How do you write an equation of a line going through #2^x-5<64#?

Sadie Alexander · Accepted Answer

As expressed this question does not make any sense.

#2^x-5 < 64# is not a point. It could be solved for #x# as #color(white)("XXX")2^x-5 < 64# #color(white)("XXX")rarr 2^x < 69# #color(white)("XXX")rarr log_2 2^x < log_2 69# #color(white)("XXX")rarr x < log_2 69# then using a calculator evaluating #log_2 69 ~~6.108524457# In the Cartesian plane this is a region composed of all points to the left of #x = 6.108524457# (approx.) I suppose you could argue that the equation #x=0# (for example) is the equation of a line going through (i.e. completely contained in) this region.

James Allen · Answer

undefined To write the equation of a line going through $2^x - 5 < 64$, we need to determine the boundary where $2^x - 5 = 64$, as well as the direction in which the line should extend.

First, solve for $x$ when $2^x - 5 = 64$:

$$2^x - 5 = 64$$
$$2^x = 69$$

Taking the logarithm of both sides to solve for $x$, we get:

$$x = \log_2(69)$$

Now, the line will extend in the direction of the inequality, which is less than. So, the line will be above the curve $2^x - 5 = 64$.

Therefore, the equation of the line going through $2^x - 5 < 64$ is simply $x = \log_2(69)$.

How do you write an equation of a line going through #2^x-5&lt;64#?

How do you write an equation of a line going through #2^x-5<64#?