Isolate the a value from this formula

Question

$$5337\cdot \left(0.1^{0.1\cdot a}+0.1^{0.2 \cdot a}+0.1^{0.3 \cdot a}\right)=10159$$

by online calculate : $a \approx 1.028222635$

I'm new at math,how to isolate the $a$ value from this formula?

Answer 1

You want : $5337\cdot(0.1^{a\cdot 0.1}+0.1^{a\cdot 0.2}+0.1^{a\cdot 0.3})=10159$

Set $x:=0.1^{a\cdot 0.1}=(0.1^{0.1})^a$ then your problem becomes : $$x+x^2+x^3=\frac{10159}{5337}$$

A third degree equation may be solved formally by WolframAlpha for example with a real solution given by :

The $a$ solutions may be obtained by using $a\cdot 0.1\ln(0.1)=\ln(x)$

and the real solution is indeed $\approx \frac{\ln(0.7891830272818)}{0.1\ln(0.1)}\approx 1.028222635581$

Answer 2

more than hint :

Substitute : $0.1^{0.1a}=b$ , so :

$$b^3+b^2+b = \frac{10159}{5337} $$

Now use general formula of roots to find values of $b$ .

Then calculate $a$ from :

$$a =10 \cdot \log_{0.1} b$$