How would one get the x-intercepts of these exponentail graphs?
y = 2 * 3^x + 1
and
y = 3^-x -1

Question

How would one get the x-intercepts of these exponentail graphs?
y = 2 * 3^x + 1
and
y = 3^-x -1

anonymous · Answer

Isolate 3^x then use log, base-3.

anonymous · Answer

uh... without a calculator?

anonymous · Answer

Yes.

anonymous · Answer

Do you want exact values or approximations?

anonymous · Answer

the exact values.

anonymous · Answer

The first one will give you problems because of taking log of a negative number, so the solution there will be complex (imaginary).

anonymous · Answer

The second one is easy because the log of 1 = 0.

campbell_st · Answer

well look at the 1st graph... an exponential graph 

y = a^x has a horizontal asymptote at x = 0

so y = a^x + 1 has the asymptote at x = 1 

so the graph will not have an x - intercept.

anonymous · Answer

That makes sense, Campbell_st. We havent been taught logarithms, Cliffsedge.

campbell_st · Answer

the 2nd graph 

y = 3^(-x) - 1 

has a horizontal  asymptote at y = -1

y = 3^{-x} would cut the y axis at y = 1

since the graph is moved down 1 the y intercept is y = 0

which means the x intercept is x = 0 with a horizontal asymptote at y = -1

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