SOLUTION: What is the approximate solution to 5^6x+3=37? This means 5 with the exponents of 5 being 6x+3 and the answer equaling 37. Any help is appreciated. Thanks, Sherry

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: What is the approximate solution to 5^6x+3=37? This means 5 with the exponents of 5 being 6x+3 and the answer equaling 37. Any help is appreciated. Thanks, Sherry      Log On


   

Question 718243: What is the approximate solution to 5^6x+3=37? This means 5 with the exponents of 5 being 6x+3 and the answer equaling 37.
Any help is appreciated.
Thanks,
Sherry
Found 2 solutions by jsmallt9, DrBeeee:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
First of all, if you put multiple-term exponents in parentheses:
5^(6x+3)=37
you don't have to explain what the exponent is.

5%5E%286x%2B3%29=37
To find a decimal approximation for the solution we will be using logarithms, specifically logs your calculator "knows" (base e, "ln", or base 10, "log):
ln%285%5E%286x%2B3%29%29=ln%2837%29
Next we use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, which allows us to move the exponent of the argument out in front (where can then "get at" the variable):
%286x%2B3%29%2Aln%285%29=ln%2837%29
Dividing by ln(5):
6x%2B3=ln%2837%29%2Fln%285%29
Subtract 3:
6x=ln%2837%29%2Fln%285%29-3
Divide by 6:
x=%28ln%2837%29%2Fln%285%29-3%29%2F6
This is an exact expression for the solution to your equation. For a decimal approximation, use your calculator. (Note: ln%2837%29%2Fln%285%29 is not the same as ln%2837%2F5%29!! So you have to find the two ln's separately and then divide, then subtract 3 and finally divide by 6.)

Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
When your variable is in an exponent, the general rule is that you use logarithms to solve for the variable. For example, if we have a power (remember your definition of a power?) equal to a number as
(1) 2^x = 3
The procedure is to take the LN of both sides of the equation and get
(2) LN(2^x) = LN(3)
Now we apply the rule that the LN of a power is equal to the exponent times the LN of the base and get
(3) x*LN(2) = LN(3)
Now you treat the LN(2) and LN(3) just as you would we any real number and get
(4) x = LN(3)/LN(2)
Now use your scienfific calculator to get
(5) x = 1.0986.../0.6931... or
(6) x = 1.584...
Rarely will it be a "nice" round number, so keep all intermediate computational values in the calculator; rounding only the final answer.
So much for the lesson, now let's do your problem
(7) 5%5E%286x%2B3%29+=+37
Note we have a power on the left and a real number on the right just as we do in example (1).
Now take the LN of (7) and get
(8) %286x%2B3%29%2ALN%285%29+=+LN%2837%29 or
(9) %286x%2B3%29+=+LN%2837%29%2FLN%285%29 or
(10) 6x+=+%28LN%2837%29%2FLN%285%29%29-3 or
(11) x+=+%28%28LN%2837%29%2FLN%285%29%29-3%29%2F6 or
(12) x = (((3.6109...)/(1.6094...))-3)/6 or
(13) x = ((2.243...)-3)/6 or
(14) x = (-0.7564...)/6 or
(15) x = -0.126...
Let's check this using (7).
Is (5^(6*(-0.126...)+3) = 37)?
Is (5^(2.243...) = 37)?
Is (37 = 37)? Yes
Answer: x is approximately -0.1260684
PS In the above calculations you could use LOG instead of LN, it doen't matter.