Solving a Logarithmic Equation with an Irrational Result

[e^(i Pi)+1=0]

Homework Statement

27^x+1=3^2x+1

The Attempt at a Solution

log(27^x+1)=log(3^2x+1)

(x+1)log(27)=(2x+1)log(3)

2x+1=$\frac{(x+1)log(27)}{log(3)}$

$\frac{2x+1}{x+1}=\frac{log(27)}{log(3)}$

x=-2

I'm wondering how I would have continued solving this for the exact answer if $\frac{log(27)}{log(3)}$ had turned out to be irrational?

 

[Curious3141]

Homework Statement

27^x+1=3^2x+1

The Attempt at a Solution

log(27^x+1)=log(3^2x+1)

(x+1)log(27)=(2x+1)log(3)

2x+1=$\frac{(x+1)log(27)}{log(3)}$

$\frac{2x+1}{x+1}=\frac{log(27)}{log(3)}$

x=-2

I'm wondering how I would have continued solving this for the exact answer if $\frac{log(27)}{log(3)}$ had turned out to be irrational?

If log(m)/log(n) was irrational (call it p), then you'd just have solved it with algebra the usual way to get x in terms of p.

i.e.

(2x+1) = px + p

x(2-p) = p-1

x = (p-1)/(2-p)

That would be an exact answer, but it wouldn't be a "nice number" (an integer or even a rational answer).

 

[cepheid]

Homework Statement

27^x+1=3^2x+1

The Attempt at a Solution

log(27^x+1)=log(3^2x+1)

(x+1)log(27)=(2x+1)log(3)

2x+1=$\frac{(x+1)log(27)}{log(3)}$

$\frac{2x+1}{x+1}=\frac{log(27)}{log(3)}$

x=-2

I'm wondering how I would have continued solving this for the exact answer if $\frac{log(27)}{log(3)}$ had turned out to be irrational?

Saying "log(a)" and "log(b)" IS exact, even if you can't express those as rational numbers.

xlog(a) + log(a) = 2x(log(b)) + log(b)

x[log(a) - 2log(b)] = log(b) - log(a)

$$x = \frac{\log(b) - \log(a)}{\log(a) - 2\log(b)}$$

and that would be your final answer, which is very much exact.

 

[e^(i Pi)+1=0]

(2x+1) = px + p

x(2-p) = p-1

I am probably being incredibly dense, but I don't follow what you did here.

 

[e^(i Pi)+1=0]

Saying "log(a)" and "log(b)" IS exact, even if you can't express those as rational numbers.

Yeah I realize that, it's like $\sqrt{2}$. The problem I'm having is with the algebra.

 

[e^(i Pi)+1=0]

Got it, I really was being dense. Thank you.

 

[RoshanBBQ]

(2x+1) = px + p

x(2-p) = p-1

I am probably being incredibly dense, but I don't follow what you did here.

$$\frac{2x+1}{x+1}=\frac{a}{b}$$
$$2x+1=\frac{a}{b}(x+1)$$
$$2x+1=\frac{a}{b}x+\frac{a}{b}$$
$$2x-\frac{a}{b}x+1=\frac{a}{b}$$
$$2x-\frac{a}{b}x=\frac{a}{b}-1$$
$$x\left (2-\frac{a}{b}\right )=\frac{a}{b}-1$$
$$x=\frac{\frac{a}{b}-1}{2-\frac{a}{b}}$$

edit: nvm, you got it