How do you solve #12=10^(x+5)-7#?

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Answer 1 · 2016-09-10T10:35:19

#x = log_(10)(19) - 5#

We have: #12 = 10^(x + 5) - 7#

Let's add #7# to both sides of the equation:

#=> 19 = 10^(x + 5)#

#=> 10^(x + 5) = 19#

Then, let's apply #log_(10)# to both sides of the equation:

#=> log_(10)(10^(x + 5)) = log_(10)(19)#

Using the laws of logarithms:

#=> (x + 5) log_(10)(10) = log_(10)(19)#

#=> (x + 5) cdot 1 = log_(10)(19)#

#=> x + 5 = log_(10)(19)#

Now, to solve for #x#, let's subtract #5# from both sides:

#=> x = log_(10)(19) - 5#