How do you solve $- 5 \cdot 10^{3 b} = - 77$ $-5*10^(3b)=-77$ ?

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Answer 1 · 2016-09-05T17:34:13

$b = \frac{1}{3} (log (77) - log (5))$ $b = (1) / (3) (log(77) - log(5))$

We have: $- 5 \cdot 10^{3 b} = - 77$ $- 5 cdot 10^(3 b) = - 77$

Let's begin by dividing both sides of the equation by $- 5$ $- 5$ :

$\Rightarrow 10^{3 b} = \frac{77}{5}$ $=> 10^(3 b) = (77) / (5)$

Then, let's apply logarithms to both sides:

$\Rightarrow log (10^{3 b}) = log (\frac{77}{5})$ $=> log(10^(3 b)) = log((77) / (5))$

Now, using the laws of logarithms:

$\Rightarrow 3 b log (10) = log (77) - log (5)$ $=> 3 b log(10) = log(77) - log(5)$

$\Rightarrow 3 b = log (77) - log (5)$ $=> 3 b = log(77) - log(5)$

$\Rightarrow b = \frac{1}{3} (log (77) - log (5))$ $=> b = (1) / (3) (log(77) - log(5))$

How do you solve -5*10^(3b)=-77−5⋅103b=−77-5*10^(3b)=-77?