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
$\begingroup$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 $b = 1,182,795,699^{\frac 1{4.89}}=71.680456772641838989153714527019$
$\endgroup$ $\begingroup$$$b^{4.89}=1 182 795 699$$ Taking natural $\log$ on both sides, $$\log b^{4.89}=\log1 182 795 699$$ $$4.89\log b=\log1 182 795 699$$ $$\log b=\frac{\log(1 182 795 699)}{4.89}$$ $$b=e^{\frac{\log(1 182 795 699)}{4.89}}$$
$\endgroup$ $\begingroup$$$b^{4.89}=1 182 795 699$$It is also equal to$$b^{489/100}=1182795699$$Hence,$$b=1182795699^{100/489}$$
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