How do I solve the equation $e^{\ln(2x+1)} = 5x$?

Question

The problem is $$e^{\ln(2x+1)} =5x$$

I've tried using natural logs to both sides like.. $2x+1= \ln 5x $ But I'm not sure if $\ln$ and $e^{\ln}$ cancel out.

Answer 1

e raised to the power of a natural log cancels out both. This just leaves you with $2x+1=5x,$ or $x=\frac{1}{3}$

Answer 2

Yes, the exponential and the logarithm are inverses: $$e^{\ln(\color{red}{\rm stuff})} = \color{red}{\rm stuff}, \quad \ln(e^{\color{blue}{\rm stuff}}) = \color{blue}{\rm stuff},$$ so that $e^{\ln(2x+1)} = 2x+1$.

Answer 3

The logarithm is the inverse of the exponential, meaning $exp \circ log = id$.

It then follows that $e^{log(2x+1)}=2x+1=5x$ which should be easy to solve.

Answer 4

$e^{ln(x)}$ $=$x

check the following pdf for this and more properties of logs

http://assets.wne.edu/95/exps_and_logs.pdf

Using this property we can see that

$2x+1$ = $5x$

$x$ $=$ ${1/3}$

Answer 5

If you apply natural log to both sides, you should instead get$$\begin{align*} \ln\left[e^{\ln(2x+1)}\right] &= \ln(5x)\\ \ln(2x+1)\cdot\ln(e) &= \ln(5x)\\ \ln(2x+1)\cdot1 &= \ln(5x)\\ \ln(2x+1) &= \ln(5x)\\ &\vdots \end{align*}$$

Answer 6

$e^{\ln a}=a$, for $a\gt-1/2$. so $2x+1=5x$ and solve for $x$.

Answer 7

Because $e^x$ and $\ln x$ both have domains $x > 0$ where they are injective then, given that they are inverses of one another means that the composition of them yields the identity function. Specifically we have $$\ln e^x = e^{\ln x} =x$$

The same logic applies to other functions, for example, consider the function $x^3$ and the function $x^{1/3}$, they are inverses of another. The cube function is the inverse of the cube root function and vice-versa. So composing them yields the identity again $$(x^3)^{1/3} = (x^{1/3})^3 = x $$

Another example would be, considering a suitable domain for $x$, that $$\arcsin \sin x = \sin \arcsin x = x$$

The list of example goes on... and I hope I've provided a valuable insight into your understanding of how composing inverses yields the identity.