1
$\begingroup$

I'm trying to solve for $b$ in what seems like a very simple exponential equation: $b^{4.89}=1 182 795 699$

I know that $y=\log_b x$ is equivalent to $b^y=x$ but I don't know how that helps me to isolate $b$. I looked for the solution in WolframAlpha and it gives me $b=71.68$, but it won't show a step-by-step solution.

I would appreciate a lot if you could tell me how to solve this equation. Thanks :)

$\endgroup$

3 Answers 3

3
$\begingroup$

When you learn logs don't forget everything you know about roots!

$b^{4.89}=1 182 795 699$

So $(b^{4.89})^{\frac 1{4.89}} = 1,182,795,699^{\frac 1{4.89}}$

So $b = 1,182,795,699^{\frac 1{4.89}}=71.680456772641838989153714527019$

$\endgroup$
2
$\begingroup$

$$b^{4.89}=1 182 795 699$$ Taking natural $\log$ on both sides, $$\log b^{4.89}=\log1 182 795 699$$ $$4.89\log b=\log1 182 795 699$$ $$\log b=\frac{\log(1 182 795 699)}{4.89}$$ $$b=e^{\frac{\log(1 182 795 699)}{4.89}}$$

$\endgroup$
0
$\begingroup$

$$b^{4.89}=1 182 795 699$$ It is also equal to $$b^{489/100}=1182795699$$ Hence, $$b=1182795699^{100/489}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .