Beginning with the function `f(x)=log_ax` , state what transformations were used on this to obtain the functions given below:

`p(x)=-5/8log_ax`

`r(x)=log_a(5-x)`

`t(x)=2log_a2x`

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log multiplied by any positive number scales it. When multiplied by a negative number, the function is scaled by that number, but also flipped about the x-axis. Therefore:

`p(x)=-(5/8)log_a(x)` is `f(x)=log_a(x)` scaled by a factor of 5/8 and flipped about the x-axis

The black line is `f(x)=log(x)` and the red line is `p(x)=(-5/8)log(x)` .

When the sign of x is negative, the function is flipped about the y-axis. When a number is added to x, the graph is translated that number of units left on the x-axis, and when a number is subtracted from x, the graph is translated that number of units to the right.

`r(x)=log_a(5-x)` ; therefore, r(x) is f(x) translated 5 units to the left and then flipped about the y-axis.

The black line is f(x) and the red line is r(x).

Finally, when you multiply x by any number, the graph is translated upwards by log(m), where m is the number x is multiplied by.

`t(x)=2log_a(x)=2log_a(2)+2log_a(x)` is `f(x)=log_a(x)` scaled by a factor of 2 and translated `log_a(2)` units upwards on the y-axis.

t(x) is the red line and f(x) is the black line.

For simplicity, all graphs were done with a=10.

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