# Solve for x logarithm question

• Ascleipus
In summary, the problem asks to solve for x, but gives no indication as to what kind of x should be used. After experimenting with different exponents, I end up with the following equation: x=ln(10) \frac{W\left(\frac{5^{5/ln(10)}}{ln(10)/4} \right)}{4} where W is Lambert's W function.
Ascleipus

## Homework Statement

The problem asks to solve for x

log(x)=-4x+5 (the log has base 10 - not natural log)

## The Attempt at a Solution

log(x)=-4x+5

10^-4x+5=x

log(10^-4x+5)=log(x)
-4x+5*log(10)=log(x)
-4x+5=log(x)

seems at first to be a simple enough question to solve, however I only end up with a circular solution bringing me back to the original question when i take the log of this to bring down the exponent (of course)

Any help would be appreciated :)

You could try writing $$log_{10}x=\frac{ln x}{ln10}$$

There isn't going to be any simple "algebraic" way to solve that because you have the unknown, x, both inside and outside the logarithm. The first thing I would do is use CAF123's suggestion to change to natural logarithm: $ln(x)/ln(10)= -4x+ 5$ so that $ln(x)= (-4/ln(10))x+ 5/ln(10)$. Now take the exponential of both sides:
$x= e^{(-4ln(10))x}e^{5/ln(10)}$

$xe^{(4/ln(10))x}= e^{5/ln(10)}$

Now let y= (4/ln(10))x so that x= (ln(10)/4)y and the equation becomes
$(ln(10)/4)ye^y= 5^{5/ln(10)}$

$$ye^y= \frac{5^{5/ln(10)}}{ln(10)/4}$$

and therefore
$$y= W\left(\frac{5^{5/ln(10)}}{ln(10)/4)}\right)$$
where "W" is Lambert's W function (http://en.wikipedia.org/wiki/Lambert_W_function) which is defined as the inverse function to $f(x)= xe^x$
(I told you there was no simple "algebraic" solution!)

Since we defined y= (4/ln(10))x, we have, finally,
$$x= ln(10)\frac{W\left(\frac{5^{5/ln(10)}}{ln(10)/4)}\right)}{4}$$
May you have joy of it!

Since this was posted under "Precalculus Mathematics", where in the world did you get this problem?

HallsofIvy said:
There isn't going to be any simple "algebraic" way to solve that because you have the unknown, x, both inside and outside the logarithm. The first thing I would do is use CAF123's suggestion to change to natural logarithm: $ln(x)/ln(10)= -4x+ 5$ so that $ln(x)= (-4/ln(10))x+ 5/ln(10)$.

Should that be ##ln(x) = ln(10)[5 - 4x]?## Instead of taking the exponential of both sides, I then said ##ln(x) = 5ln(10) - 4x ln(10)\,\Rightarrow ln(x) = ln \left(\frac{10^5}{10^{4x}} \right)##. I now take exp to get ##x10^{4x} = 10^5.##

it will be easier to solve this question using Newton-Raphson method. :)

CAF123 said:
Should that be ##ln(x) = ln(10)[5 - 4x]?## Instead of taking the exponential of both sides, I then said ##ln(x) = 5ln(10) - 4x ln(10)\,\Rightarrow ln(x) = ln \left(\frac{10^5}{10^{4x}} \right)##. I now take exp to get ##x10^{4x} = 10^5.##
Yes, I messed that up. Thanks for catching it. But that still has no "algebraic" solution. In order to use the W function, you will need to convert to "e". $$10^{4x}= e^{ln(10^{4x}}= e^{4xln(10)}$$ so the equation is the same as $$xe^{4xln(10)}= 10^5$$.

Let y= 4x ln(10) so that x= y/(4 ln(10)). The equation becomes, in terms of y, $$\frac{ye^y}{4ln(10}= 10^5$$. $$ye^y= 400000 ln(10)$$, $$y= W\left(40000 ln(10)\right)$$.

Then $$x= \frac{W\left(40000 ln(10)\right)}{4 ln(10)}$$

Ascleipus said:

## Homework Statement

The problem asks to solve for x

log(x)=-4x+5 (the log has base 10 - not natural log)

You can solve the equation numerically by iteration.
Rewrite it as

x=(5-log(x))/4,

Get the next approximation from a previous one by xk+1=(5-log(xk))/4.

Using x1=1, substitute and get the next approximation: x2=1.25. Substitute again, now you get x3=1.227 ...

ehild said:
You can solve the equation numerically by iteration.
Rewrite it as

x=(5-log(x))/4,

Get the next approximation from a previous one by xk+1=(5-log(xk))/4.

Using x1=1, substitute and get the next approximation: x2=1.25. Substitute again, now you get x3=1.227 ...
Well, sure, if you want to do it the easy way!

## What is a logarithm?

A logarithm is a mathematical function that helps us solve exponential equations. It tells us what power we need to raise a base number to in order to get a specific number.

## What does it mean to "solve for x" in a logarithm question?

Solving for x in a logarithm question means finding the value of x that makes the equation true. In other words, you are trying to find the number that you need to raise the base to in order to get the given number.

## How do I solve a logarithm equation?

To solve a logarithm equation, you need to use the properties of logarithms to simplify the equation. Then, you can use algebraic techniques to isolate the variable, x, on one side of the equation. Finally, you can use the inverse property of logarithms to find the value of x.

## What are the common properties of logarithms?

The common properties of logarithms are the product property, quotient property, power property, and inverse property. These properties help us simplify logarithm equations and solve for x.

## Can a logarithm have more than one solution?

Yes, a logarithm can have more than one solution. This is because the exponential function is not a one-to-one function, meaning that different values of x can result in the same value of the logarithm. However, in most cases, we are only interested in the principal solution, which is the positive value of x that satisfies the equation.

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