Why is y=In x the only logarithmic function with a gradient 1 at x=1?

Question

I know nothing about logarithmic functions, only they are supposed to be inverse to exponentials?

I just need a written out explanation to why only natural logarithmic functions have a gradient of 1 at x-intercept=1.

Thank you!

Answer 1

Regarding your question, why is y = ln(x) the only logarithmic function with a gradient of 1 at x intercept?

Whenever we talk about gradients, we generally talk about derivatives at that point. So lets differentiate a more general logarithm, $y = log_a(x)$. You can look at many sources for the proof of the derivative.

$$y = log_a(x)$$ $$y' = \frac{1}{ln(a)x}$$

Notice at we have the function ln(x) which have base e (a = e). So the derivative will be $y' = \frac{1}{ln(e)x} = \frac{1}{x}$ becaue ln(e) = 1 by definition of the logarithm function.

Let's find the gradient at x intercept (x, 0). Find what x is. $$y = ln(x)$$ $$0 = ln(x)$$ $$1 = x$$

The x intercept of y = ln(x) is when x = 1.

So now we just have to find the gradient at x = 1. $$y' = \frac{1}x$$ at x = 1 $$y' = \frac{1}1 = 1$$ So in conclusion, the reason why is y = ln(x) the only logarithmic function with a gradient of 1 at x interecept, is because ln(a) = 1. As seen at the above. if a is not e, then ln(a) will not be equal to 1, thus at x intercept, the derivative is not 1.

Answer 2

I assume you mean $y = \ln(x)$ where $ln$ is the logarithm with base $e$. We can write any logarithm like this $$\log_a(x) = \frac{\ln(x)}{\ln(a)}$$ This means that $$\frac{d}{dx} (\log_a(x))= \frac{1}{x\ln(a)}$$ The x-intercept occurs when $y= 0 \implies 0 = \log_a(x) \implies a^0 = x \implies 1 =x$. Hence the gradient at the x-interception will be $$\frac{1}{1\ln(a)}=\frac{1}{\ln(a)}$$ If this is equal to one then $$1 = \frac{1}{\ln(a)} \implies \ln(a) = 1 \implies a = e^1 = e$$ Therefore only the logarithm base $e$ has a gradient of one at the intercept