exponential and log functions

find the x-coordinate of the point where the graph of the curve with the equation y=log3(x-2)+1 intersects with the x axis.

Any help would be great.

Reply 1

DaveManUK

solve for y = 0

Reply 2

Newton

f(x)=log_3_(x-2)+1

The x-intersect =>f(x)=0

=>log_3_(x-2)=-1

Raise each side to 3

=>(x-2)=(3^(-1))

=>x-2=(1/3)

=>3x-6=1

=>3x=7

=>x=(7/3)

Newton.

Reply 3

Christophicus

kay50

find the x-coordinate of the point where the graph of the curve with the equation y=log3(x-2)+1 intersects with the x axis.

Any help would be great.

At x-axis, y = 0
log3(x-2) + 1 = 0
log3(x-2) = -1

Let (x-2) = y
log3y = -1
log10y/log103 = -1 (using the change of base formula)

www.uncwil.edu/courses/mat111hb/EandL/logprop/base.gif

log10y = log10(1/3)

Therefore y = 1/3
But, y = (x-2)
Therefore;
x-2 = 1/3
x = 2 + 1/3 = 7/3