# How can I solve the problem of $e^x-\ln{x}=4$ without a calculator?

• MHB
• karush
In summary: Specifically, if the problem is to solve f(x)= 0 and we choose $x_0$ as our starting point, the tangent line is $y= f'(x_0)(x- x_0)+ f(x_0)$. Setting that equal to 0, $f'(x_0)(x- x_0)+ f(x_0)= 0$ so $f'(x_0)(x- x_0)= -f(x_0)$. $x- x_0= -\frac{f(x_0)}{f'(x_0)}$ and then $x= x_0- \frac{ karush Gold Member MHB solve$e^x-\ln{x}=4\quad
x\approx1.48
\quad x\approx 005$ok I could only do this with a calculator but need steps Last edited: Newton's method solves this readily enough. Yes, either use a calculator or do a whole lot of arithmetic with paper and pencil! To answer your question, even though it's easy to show that a root exists, it's actually impossible to get an exact form for. As others have suggested, if you want to get an approximate solution, you'll need an iterative method such as Newton's Method, or get the CAS to solve it for you. romsek said: Newton's method solves this readily enough. don't think I would know how to use Newtons method on this Prove It said: To answer your question, even though it's easy to show that a root exists, it's actually impossible to get an exact form for. As others have suggested, if you want to get an approximate solution, you'll need an iterative method such as Newton's Method, or get the CAS to solve it for you. I quit buying handheld calculators I had a TI inspire CS CAS but the keyboard wore out W|A can calculate it karush said: I quit buying handheld calculators I had a TI inspire CS CAS but the keyboard wore out W|A can calculate it Which is a CAS... Prove It said: Which is a CAS... yes but its not a handheld that requires e charging and is easily lost and a hard to read screen To use Newton's method, we select some starting point, hopefully close to a root of the equation but not necessarily. If that point happens to be a root, we are done. If not then we construct the tangent line to the graph of the function at that point and solve the linear equation to see where that tangent line crosses the x-axis. Hopefully, if that new x value is not a root, it is closer so we do it again. Specifically, if the problem is to solve f(x)= 0 and we choose$x_0$as our starting point, the tangent line is$y= f'(x_0)(x- x_0)+ f(x_0)$. Setting that equal to 0,$f'(x_0)(x- x_0)+ f(x_0)= 0$so$f'(x_0)(x- x_0)= -f(x_0)$,$x- x_0= -\frac{f(x_0)}{f'(x_0)}$and then$x= x_0- \frac{f(x_0)}{f'(x_0)}$. THAT is "Newton's method"-to solve f(x)= 0, select some starting$x_0$and calculate$x_1= x_0- \frac{f(x_0)}{f'(x_0)}$. Then calculate$x_2= x_1- \frac{f(x_1)}{f'(x_1)}$and keep repeating that until you get the accuracy you want. Typically you know you are close to the desired root because the values are close together. For example, if you want the root "correct to three decimal places" keep repeating until successive values of x are correct to three decimal places. Here,$f(x)= e^x- log(x)- 4$and$f'(x)= e^x-\frac{1}{x}$so that$x_1= x_0- \frac{f(x_0)}{f'(x_0)}= x_0-\frac{e^{x_0}- log(x_0)- 4}{e^{x_0}- \frac{1}{x_0}}$. Choose some starting value, say$x_0=1$or$x_0= 2\$ and start calculating!

If you were as old as I am you would have learned to use tables to find the exponential and logarithm but for God's sake use a calculator!

Mahalo that was a great help

## 1. What is the meaning of "-6.6.72 log and e problem"?

The "-6.6.72 log and e problem" refers to a mathematical equation involving the logarithm and the mathematical constant e. The equation typically looks like -6.6.72 log e, where the numbers and symbols in between the logarithm and e can vary.

## 2. How do you solve the "-6.6.72 log and e problem"?

The "-6.6.72 log and e problem" can be solved by using the properties of logarithms and the value of e, which is approximately 2.71828. Depending on the specific equation, you may need to use a calculator or solve it algebraically.

## 3. What is the purpose of using logarithms and e in this problem?

Logarithms and e are commonly used in mathematical equations to help solve for unknown variables or simplify complex expressions. They have many applications in fields such as science, engineering, and finance.

## 4. Can the "-6.6.72 log and e problem" have multiple solutions?

Yes, the "-6.6.72 log and e problem" can have multiple solutions depending on the specific equation and the given values. It is important to carefully check your work and consider any restrictions on the variables to ensure you have found all possible solutions.

## 5. Is there a real-life application for the "-6.6.72 log and e problem"?

Yes, there are many real-life applications for the "-6.6.72 log and e problem" in fields such as finance, physics, and biology. For example, logarithms and e are used in calculating interest rates, radioactive decay, and population growth.

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