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secondary 3 | A Maths
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my wrong solution:
ln(x) =-2x
e^-2x=x
y=x
Thus there's 0 solution as lines do not intersect
However, the answer sheet says there's 2 solutions. What went wrong in my steps?
Date Posted: 2 years ago
Views: 346
There should only be one solution.
y = e^-2x is actually a reflection of the graph of y = e^2x in the vertical axis, so it’s a going down graph.
On the other end, y = x is an upward sloping line.
They are going to meet only once.
I also considered the original equation ln x = -2x, where y = ln x is an upward graph and y = -2x is a downsloping line, with one intersection.
y = e^-2x is actually a reflection of the graph of y = e^2x in the vertical axis, so it’s a going down graph.
On the other end, y = x is an upward sloping line.
They are going to meet only once.
I also considered the original equation ln x = -2x, where y = ln x is an upward graph and y = -2x is a downsloping line, with one intersection.
ohhh I see.. thank you so much!! however, may I know what your graphs looks like as a whole? I have uploaded mine as one of the answers under this question, but it seems wrong :( I'm not sure what is the correct graph
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Left is what I plotted and right is the photomath graph... Why are they different??
Date Posted:
2 years ago
When equations are transformed, they result in their corresponding graphs being different, of course! What remains unchanged as we transform the equations is the number of solutions.
Take a simple case of trying to solve the inequality x^2 - 2x > 3.
We plot the graph of y = x^2 - 2x and see where the curve lies above the line y = 3.
Moving the 3 to the other side results in x^2 - 2x - 3 > 0. If we plot the new graph of y = x^2 - 2x - 3, the curves are going to be in entirely different position. However, this time we are solving for y > 0, and the resulting range of values for x remains unchanged.
Take a simple case of trying to solve the inequality x^2 - 2x > 3.
We plot the graph of y = x^2 - 2x and see where the curve lies above the line y = 3.
Moving the 3 to the other side results in x^2 - 2x - 3 > 0. If we plot the new graph of y = x^2 - 2x - 3, the curves are going to be in entirely different position. However, this time we are solving for y > 0, and the resulting range of values for x remains unchanged.
The left graph is the graph of the transformed equations: y = e^-2x versus y = x.
The right graph is the graph of the original equations: y = ln x against y = -2x.
The right graph is the graph of the original equations: y = ln x against y = -2x.
In other words, whenever we try to modify an equation (such as while trying to transform a long expression into a quadratic expression), we are changing the dynamics of the graphs if we were to plot them on paper.
But the solutions to the equation remain unchanged even as we modify our equations to obtain x.
But the solutions to the equation remain unchanged even as we modify our equations to obtain x.
I got it!! Thank you so much
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