Rearranging Logarithms: Finding the Solution to log Equations

[OnlinePhysicsTutor]

Homework Statement

2 - log₁₀ 3^x = log₁₀(x/12)

Homework Equations

loga^b=b log a
log(a/b)= log a - log b

The Attempt at a Solution

2 + log₁₀ 12= log₁₀ x - x log₁₀ 3
Start seems simple but cannot see where to go from here, taking exponentials doesn't seem to help. Not sure what the next steps could be.

 

[Buffu]

log is base 10 or it is natural log ?

 

[OnlinePhysicsTutor]

I assume it is meant to be base 10, so I have edited post to include the base.

 

[Buffu]

I assume it is meant to be base 10, so I have edited post to include the base.

After rearranging a bit I can't see how this has a nice solution.

I get $5^2 *2^4*3 = x3^x$.

 

[OnlinePhysicsTutor]

After rearranging a bit I can't see how this has a nice solution.

I get $5^2 *2^4*3 = x3^x$.

I'm happy with that as well, thanks.

 

[Mark44]

Homework Statement

2 - log₁₀ 3^x = log₁₀(x/12)

Homework Equations

loga^b=b log a
log(a/b)= log a - log b

The Attempt at a Solution

2 + log₁₀ 12= log₁₀ x - x log₁₀ 3
Start seems simple but cannot see where to go from here, taking exponentials doesn't seem to help. Not sure what the next steps could be.

Your equation is equivalent to $100 = \frac x {12} \cdot 3^x$. Because the variable occurs both as an exponent and as a multiplier, there are not any simple analytic ways to solve this equation. However, you can get good approximations by numeric means, simply by substituting value for x on the right side, and comparing the result with 100 on the left side. Using a spreadsheet I see that there is a solution near x = 50.16. The actual solution is slightly smaller than this.

Edit: As Ray points out, my number here is incorrect. I was using an incorrect formula in my spreadsheet, using log(3^x) instead of 3^x.

 

[Ray Vickson]

Your equation is equivalent to $100 = \frac x {12} \cdot 3^x$. Because the variable occurs both as an exponent and as a multiplier, there are not any simple analytic ways to solve this equation. However, you can get good approximations by numeric means, simply by substituting value for x on the right side, and comparing the result with 100 on the left side. Using a spreadsheet I see that there is a solution near x = 50.16. The actual solution is slightly smaller than this.

For If $f(x) = (x/12) 3^x,$ we have$f(50) = (50/3) e^{50} \doteq 0.299 \times 10^{25}$, so the solution of $f(x) = 100$ must certainly be quite a bit less than 50. Maple gets $x \doteq 4.990.$

 

[Buffu]

For If $f(x) = (x/12) 3^x,$ we have$f(50) = (50/3) e^{50} \doteq 0.299 \times 10^{25}$, so the solution of $f(x) = 100$ must certainly be quite a bit less than 50. Maple gets $x \doteq 4.990.$

what is $\doteq$ ?

 

[Ray Vickson]

what is $\doteq$ ?

$\doteq$ means "approximately equal to", sometimes also written as $\approx$. I avoid using "=" in such cases just so the reader will understand that the answer is not exactly 4.990. For example, a better approximation is obtained by using 60 digits of precision, giving
$x \doteq 4.99043541467729841484302401855197675632523233638262678465047$ Even that is not exact.

 

[Buffu]

As a additional exercise, Can we prove that there is no nice real solution for equation, Also can we know the nature of the solution ?

 

[Mark44]

For If $f(x) = (x/12) 3^x,$ we have$f(50) = (50/3) e^{50} \doteq 0.299 \times 10^{25}$, so the solution of $f(x) = 100$ must certainly be quite a bit less than 50. Maple gets $x \doteq 4.990.$

You are correct. Somehow I mistakenly had log(3^x) in my spreadsheet formula, not log(3^x) as it should have been.