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

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Answer 1 · 2016-09-05T17:34:13

#b = (1) / (3) (log(77) - log(5))#

We have: #- 5 cdot 10^(3 b) = - 77#

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

#=> 10^(3 b) = (77) / (5)#

Then, let's apply logarithms to both sides:

#=> log(10^(3 b)) = log((77) / (5))#

Now, using the laws of logarithms:

#=> 3 b log(10) = log(77) - log(5)#

#=> 3 b = log(77) - log(5)#

#=> b = (1) / (3) (log(77) - log(5))#