I am having a logarithmic function of the form:
$\dfrac{\log(1+0.35\alpha)}{\log(1+\alpha)}=0.8\;.$
I want to find out the value of alpha that solves this equation. For that I am taking log term on the denominator to the other side.
$\log(1+0.35\alpha)=0.8\log(1+\alpha)\;.$
Since base of logarithm is 10,the equation becomes
$1+0.35\alpha=(1+\alpha)^{0.8}\;.$
But since one term is having power $0.8$, how to solve this. Is there any alternative method for solving this type of logarithmic equations.
I suppose that graphing, you noticed that the solution is a large number.
To get an estimate, "ignore" the $1$'s
$$\log \left(0+\frac{ 7}{20}a\right)=\frac 4 5 \log \left(0+a\right)\implies \frac{7 }{20}a=a^{\frac 4 5 }\implies a_0=\frac{3200000}{16807}$$ Now, consider that you look for the zero of function $$f(a)=\log \left(1+\frac{7 }{20}a\right)-\frac 4 5 \log \left(1+a\right)$$ and compute the first iteration of Newton method using the previous $a_0$.
It gives $$a_1=a_0-\frac{\log \left(1+\frac{7 }{20}a_0\right)-\frac 4 5 \log \left(1+a\right) } {\frac{7 a_0-45}{5 (a_0+1) (7 a_0+20)}}$$ that is to say $$a_1=\frac{3200000}{16807}-\frac{10448253472140 }{10393297537}\left(\log \left(\frac{162401}{2401}\right)-\frac{4}{5} \log \left(\frac{3216807}{16807}\right)\right)$$
Converted to decimals, $a_0=190.397$, $a_1=179.636$ while the "exact" solution is $179.919$