Solve for x logarithm question

[Ascleipus]

Homework Statement

The problem asks to solve for x

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

Homework Equations

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 :)

 

[CAF123]

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

 

[HallsofIvy]

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?

 

[CAF123]

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 $x10^{4x} = 10^5.$

 

[thchian]

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

 

[HallsofIvy]

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

 

[ehild]

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 x_k+1=(5-log(x_k))/4.

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

 

[HallsofIvy]

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 x_k+1=(5-log(x_k))/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!