Solve a = b^(logx), for x: Steps & Explanation

[astonmartin]

I can't seem to remember how you solve a = b^(logx), for x. What are the steps for simplifying this again?

 

[Unrest]

Remove the functions around x by applying their inverse functions to both sides.

So first take log_b of both sides, then exponentiate them.

 

[HallsofIvy]

I would NOT recommend taking log base b. Instead take log of whatever base that is in your equation.

 

[Unrest]

I would NOT recommend taking log base b. Instead take log of whatever base that is in your equation.

Any reason for that? Seems like it would just make things confusing. What's wrong with:

a = b^[log(x)]
log_b(a) = log(x)
exp[log_b(a)] = x

Here I assume log(x) means natural log. Otherwise replace exp() with 10^ or whatever the base is.

 

[hradecek]

$$a&=b^\log{x}$$
$$\log{a}&=\log{b}\log{x}$$
$$\log{x}&=\frac{\log{a}}{\log{b}}$$
$$\log{x}&=\log_b{a}$$

and from definition of logarithm:
$$x=10^\log_b{a}$$

 

[Unrest]

$$a&=b^\log{x}$$
$$\log{a}&=\log{b}\log{x}$$
$$\log{x}&=\frac{\log{a}}{\log{b}}$$
$$\log{x}&=\log_b{a}$$

and from definition of logarithm:
$$x=10^\log_b{a}$$

Isn't that just the same, but with unnecessary extra steps requiring 2 additional log rules?

 

[HallsofIvy]

I just don't see any point in messing with the different bases.

From $y= b^{log(x)}$, we get [itex]log(y)= log(b)log(x)[itex]so that <br /> $$log(x)= \frac{log(y)}{log(b)}$$<br /> <br /> $$x= e^{\frac{log(y)}{log(b)}}$$<br /> or <br /> $$x= 10^{\frac{log(y)}{log(b)}}$$[/itex][/itex]

 

[Unrest]

I just don't see any point in messing with the different bases.

It depends what your intention is. If you feel uglied having logs with strange bases in the intermediate working then better do it your way.

My way has the advantage of seperating out two tasks -
1. Make x the subject (main goal)
2. Get the logs into a base you can evaluate easily (optional extra)

Maybe it's my programmer mentality trying to avoid 'spaghetti maths'.