1. The graph of y=2logbase5X-7 is reflected in the x-axis and translated 6 units up. I got y=-2logbase5X-1 as a result. The answer key says y=-2logbase5+13. How????
assuming you mean y=2log_5(x)-7
reflected it is y = -2log_5(x)+7
shifting up you just add 6:
y = -2log_5(x)+13
The graph of y= log(6(X-3)) is horizontally expanded by a factor of 3, and translated 9 units left.
I don't get that one too^
To understand how to arrive at the correct answer, let's break down the given transformation step by step:
1. Start with the original equation: y = 2 log₅(x) - 7.
2. Reflecting in the x-axis: Remember that reflecting a graph in the x-axis changes the sign of the y-values. So, by reflecting the equation, the negative sign should be applied to the entire equation.
Reflected equation: y = -2 log₅(x) + 7.
3. Translating 6 units up: To shift a graph up, we need to add the specified amount to the y-values. In this case, we need to add 6 to the equation obtained from the previous step.
Translated equation: y = -2 log₅(x) + 7 + 6.
4. Simplify the equation: Combine the constants.
Final equation: y = -2 log₅(x) + 13.
Therefore, the correct equation after reflecting in the x-axis and translating 6 units up is y = -2 log₅(x) + 13.
Note: It seems that you made an error in the last term of your solution. Instead of y = -2 log₅(x) - 1, it should be y = -2 log₅(x) + 13.