If the graph of ln y against ln x is a straight line with gradient 4 and y-intercept 6, find the relationship between x and y.
Here's my approach:
$$ln y = ln x$$
$$e^{lnx} = y$$
However,
$$y = 4x + 6$$
...from the information provided in the question.
Therefore,
$$e^{lnx} = 4x + 6$$
$$lnx = ln (4x+6)$$
$$4x + 6 = x$$
$$3x+6 = 0$$
And this is where I get stuck.
I have a feeling that my approach from the beginning is flawed, but cannot think of another way to interpret the question.
The answer is the following:
$$y = e^6x^4$$
Any help will be much appreciated, thanks in advance.
$\endgroup$ 21 Answer
$\begingroup$$\log(y) = 4\log(x) + 6$. Therefore, $\log(y) = \log x^{4} + 6$. From this we get $\log(y/x^{4}) = 6$ From which we get $y = e^{6}x^{4}$
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