SOLUTION: im so lost with this one... Solve the exponential equation. express the solution set in terms of natural logarithms. 5^3^x=3.5

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Question 622749: im so lost with this one...
Solve the exponential equation. express the solution set in terms of natural logarithms.
5^3^x=3.5
Found 2 solutions by Alan3354, jsmallt9:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the exponential equation. express the solution set in terms of natural logarithms.
5^3^x=3.5
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If you mean 5%5E%283%5Ex%29+=+3.5
3%5Ex%2Aln%285%29+=+ln%283.5%29
3%5Ex+=+ln%283.5%29%2Fln%285%29
x%2Aln%283%29+=+ln%28ln%283.5%29%2Fln%285%29%29
x+=+ln%28ln%283.5%29%2Fln%285%29%29%2Fln%283%29

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
5%5E3%5Ex=3.5
When a variable is in an exponent we often use logarithms to solve the equation. The fact that the variable is in the exponent of an exponent only means that we will go through two rounds of logarithms.

The base of logarithm we use does not matter really. But there are advantages to choosing certain bases:
  • Matching the base of the logarithm to the base of the exponent will result in a simpler expression for the answer.
  • Choosing a base your calculator "knows", log or ln, will result in an expression that we can more easily turn into a decimal approximation.
I will do the problem using both choices so you can see them both.

Matching the bases.
I'll start by finding the base 5 log of each side:
log%285%2C+%285%5E3%5Ex%29%29=log%285%2C+%283.5%29%29
Now we use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, to "move" the exponent of the argument out in front:
3%5Ex%2Alog%285%2C+%285%29%29=log%285%2C+%283.5%29%29
By definition log%285%2C+%285%29%29+=+1 (which is why matching bases results in simpler expressions) this becomes:
3%5Ex=log%285%2C+%283.5%29%29
Now we'll use base 3 logs (matching bases again):
log%283%2C+%283%5Ex%29%29+=+log%283%2C+%28log%285%2C+%283.5%29%29%29%29
Using the property to move the exponent again:
x%2Alog%283%2C+%283%29%29+=+log%283%2C+%28log%285%2C+%283.5%29%29%29%29
which simplifies to:
x+=+log%283%2C+%28log%285%2C+%283.5%29%29%29%29
This is the simplest exact expression for the solution to your equation.

Using log or ln.
Since the steps are mostly the same, I'll leave out the commentary except to explain differences.
5%5E3%5Ex=3.5
ln%285%5E3%5Ex%29=ln%283.5%29
3%5Ex%2Aln%285%29=ln%283.5%29
ln(5) does not disappear like log%285%2C+%285%29%29 did. Dividing both sides by ln(5):
3%5Ex=ln%283.5%29%2Fln%285%29
ln%283%5Ex%29=ln%28ln%283.5%29%2Fln%285%29%29
x%2Aln%283%29=ln%28ln%283.5%29%2Fln%285%29%29
ln(3) does not disappear like log%283%2C+%283%29%29 did. Dividing both sides by ln(3):
x=ln%28ln%283.5%29%2Fln%285%29%29%2Fln%283%29
While not as simple as our earlier expression, this is also an exact expression for the answer. And this one is more easily converted to a decimal approximation. (Approximately -0.228 if you're curious.)