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How do you solve #3^(x+7)=4^x#?
1 Answer
#x = 7/(log_3 4 - 1)#
or, alternatively,
#x = -7ln3/(ln(3/4)#
Explanation:
When I see these types of problems, it's helpful for me to rewrite the equation such that the bases of the two exponentials are the same. What do I mean by this?
Well, the logarithm and the exponential are inverse functions, right? So, it's logical that 4 is actually the same thing as #3^(log_3 4)#.
Using this logic we can write the original equation as
Lingon X 7 4 3 X 4
#3^(x + 7) = (3^(log_3 4))^x#
And now, using some laws of exponentials, we can simplify the above equation as
#3^(x + 7) = 3^(x*log_3 4)#
Now, how does this help us? Well, we can now take the base-3 logarithm of both sides of the equation:
#log_3 3^(x + 7) = log_3 3^(x*log_3 4)#
The logarithms will cancel with the exponentials, leaving us with
#x + 7 = x*log_3 4#
From here, we just need some simple algebra to solve for #x#:
#7 = xlog_3 4 - x#
Factor #x#:
#7 = x(log_3 4 - 1)#
And then divide:
Lingon X 7 4 3 X 4
#3^(x + 7) = (3^(log_3 4))^x#
And now, using some laws of exponentials, we can simplify the above equation as
#3^(x + 7) = 3^(x*log_3 4)#
Now, how does this help us? Well, we can now take the base-3 logarithm of both sides of the equation:
#log_3 3^(x + 7) = log_3 3^(x*log_3 4)#
The logarithms will cancel with the exponentials, leaving us with
#x + 7 = x*log_3 4#
From here, we just need some simple algebra to solve for #x#:
#7 = xlog_3 4 - x#
Factor #x#:
#7 = x(log_3 4 - 1)#
And then divide:
#7/(log_3 4 - 1) = x#
And there is our answer.
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