The graph of y = f (x) = b^x, where b > 1, is translated such that the equation of

Question

the new graph is expressed as
y – 2 = f (x – 1). The range of the new function is

A. y > 2
B. y > 3
C. y > –1
D. y > –2

2)If log7 m = log 2/3 , then the value of m, correct to the nearest hundredth

i am really confused in these questions..help please!

thank u in advance

Answer 1

the graph of y = b^x , b > 0, has the x-axis as a horizontal asymptote and its range is y > 0
rewriting your function
as
y = f (x – 1) + 2
we can see that the new function has been moved UP two units, so its range is
y > 2

By log7 m = log 2/3
do you mean
log ₇ m = log (2/3).

I will wait til your clarify?

Answer 2

Hi,

for question 1, I got y>2 as well but the answer is y>3 in my book.

and yes for question 2

Answer 3

trust yourself, you have "raised" the graph of the function by 2

2.
general rule: log _ab = log b/log a

so log m/log 7 = log2 - log3
log m = log7(log2-log3)
log m = -.1488
m = 10^-.1488
m = .70988

Answer 4

thank u =)