SOLUTION: Solve the exponential equation and express approximate solutions to the nearest hundredth. 7^x + 8 = 38

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Question 83403: Solve the exponential equation and express approximate solutions to the nearest hundredth.
7^x + 8 = 38
Found 2 solutions by rapaljer, bucky:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
I am assuing that you probably mean this for the equation:
7%5E%28x+%2B+8%29+=+38

Take the ln or the log of each side, whichever you prefer.
ln+7%5E%28x%2B8%29+=+ln+38+

By law of logarithms,
%28x%2B8%29+ln+7+=+ln+38+

By distributive property:
x+ln+7+%2B+8+ln+7+=+ln+38+
x+ln+7+=+ln+38+-+8+ln+7+

Divide by ln 7:
x=+%28ln+38+-+8+ln+7%29%2F%28ln+7%29+

Using a calculator, you should have
x=+-6.1306 which of course rounds off to x=-6.13

You may want to see my Lesson Plans on Logarithms in algebra.com. Also, check out my website by clicking on my tutor name "rapaljer" anywhere in algebra.com, and look for "MATH IN LIVING COLOR", then look in College Algebra. My entire Chapter 4 of College Algebra is devoted to LOGARITHMS.

R^2 at SCC

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
7%5Ex+%2B+8+=+38
.
solve for x
.
Isolate the term containing x on the left side and have everything else on the other side.
You can do this by eliminating +8 from the left side. This is done by subtracting 8 from
both sides to get:
.
7%5Ex+=+30
.
Now you can use the property of logarithms that says:
.
log%28a%5Eb%29+=+b%2Alog%28a%29
.
Take the logarithm of both sides:
.
log%287%5Ex%29+=+log%2830%29
.
By using the property given above, we convert the left side of the logarithm equation to:
.
x%2Alog%287%29+=+log%2830%29
.
Now solve for x by dividing both sides by log(7). This division causes the equation
to become:
.
x+=+%28log%2830%29%29%2F%28log%287%29%29
.
Note that log(30) and log(7) are just numbers. If these logs are to the base 10 or to the
base e (base e logs are commonly identified as ln), you can get them from a scientific
calculator. Putting 30 into your calculator and pressing the log key will show you that
in base 10 the log of 30 is 1.477121255. Similarly in base 10 the log of 7 is 0.84509804.
Substituting these values into the equation for x results in:
.
x+=+log%2830%29%2Flog%287%29+=+1.477121255%2F0.84509804+=+1.747869697
.
Rounding this answer to the nearest hundredth gives:
.
x+=+1.75
.
Let's check to see if this value of x makes the original given equation true. Start by
substituting 1.75 for x and this makes the left side of original equation become:
.
7%5E1.75+%2B+8
.
A calculator can be used to find that 7%5E1.75 is 30.12461949
.
When this is substituted for 7%5E1.75 the left side of the original equation becomes:
.
30.12461949+%2B+8+=+38.12461949
.
This is approximately the 38 that the original problem says it should be. The difference
is caused by rounding x to the nearest hundredth instead of carrying it to more decimal
places.
.
Hope this helps you to understand a property of logarithms that can be used when a variable
appears as an exponent.