SOLUTION: Solve the following exponential equations using Logarithms. Round to 3 decimal places. Please show work, I'm very confused. 2^x = 30 5^(x

SOLUTION: Solve the following exponential equations using Logarithms. Round to 3 decimal places. Please show work, I'm very confused. 2^x = 30 5^(x - 1) = 3x 3.53^(x + 1) = 65.4

Question 478470: Solve the following exponential equations using Logarithms.
Round to 3 decimal places. Please show work, I'm very confused.
2^x = 30
5^(x - 1) = 3x
3.53^(x + 1) = 65.4
16^(x - 4) = 3^(3 - x)
7^(x - 2) = 5^x

Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
2^x = 30
Take the log of both sides to get:
x(log(2)) = log(30)
x = [log(30)]/[log(2)]
---------------------------------
5^(x - 1) = 3x
Graph and left side.
Graph the right side.
Find any points of intersection.
----------------------------------
3.53^(x + 1) = 65.4
(x+1)log(3.53) = log 65.4
x+1 = [log(65.4)]/[log(3.53)]
etc.
----------------------------------
16^(x - 4) = 3^(3 - x)
(x-4)log(16) = (3-x)log(3)
[log(16)+log(3)]x = 4log(16)+3log(3)
Solve for "x"
---------------------------------
7^(x - 2) = 5^x
Same process as above.
----
Cheers,
Stan H.

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
exponential equations are equations that have the unknown variable in the exponent.
they are solved using logarithms.
the general form is:
b^x = c
you take the log of both sides of the equation to get:
log(b^x) = log(c)
you use the laws of logarithms as necessary to transform the equation into a form that can be solved easily.
in this case, the law used is log(b^x) = x*log(b)
your equation becomes:
x*log(b) = log(c)
you divide both sides of this equation by log(b) to get:
x = log(c)/log(b)
since b and c are known (x is the only unknown), you use your calculator to solve.
you can take the log of both sides of the equation (LOG function on your calculator) or you can take the natural log of both sides of the equation (LN function on your calculator).
either one will solve the problem for you.
these are the basic laws of logarithms that you will use:
if a = b then log(a) = log(b)
log(a*b) = log(a) + log(b)
log(a/b) = log(a) - log(b)
log(a^b) = b*log(a)
you also need to get all the unknown value on the same side of the equation before taking the log of both sides of the equation.
in doing this, you should divide or multiply both sides of the equation rather than add or subtract.
an example:
16^(x - 4) = 3^(3 - x)
divide both sides of the equation by 3^(3-x) rather than subtracting 3^(3-x) from both sides of the equation.
see the solution for this problem down below to see how this was done.
your problems are:
-----
2^x = 30
take log of both sides to get:
log(2^2) = log(30) which becomes:
x*log(2) = log(30) which becomes:
x = log(30)/log(2)
use your calculator to get:
x = 4.906890596
plug into your original equation to get:
2^4.906890596 = 30
solve to get:
30 = 30, confirming the value of x is good.
-----
5^(x - 1) = 3x
?????
x is not in the exponent which causes difficulties.
check the bottom after all other problems have been solved to see what i mean.
-----
3.53^(x + 1) = 65.4
take log of both sides of the equation to get:
log(3.53^(x+1)) = log(65.4) which becomes:
(x+1) * log(3.53) = log(65.4) which becomes:
x+1 = log(65.4)/log(3.53) which becomes:
x = log(65.4)/log(3.53)-1 which beomes:
x = 2.314460727
plug into your original equation to get:
3.53^(2.314460727 + 1) = 65.4
plug into your calculator to get:
65.4 = 65.4, confirming the value of x is good.
-----
16^(x - 4) = 3^(3 - x)
take log of both sides of the equation to get:
log(16^(x-4) = log(3^(3-x)) which becomes:
(x-4)*log(16) = (3-x)*log(3) which becomes:
(x-4)/(3-x) = log(3)/log(16) which becomes:
(x-4)/(3-x) = .396240625
multiply both sides of the equation by (3-x) to get:
x-4 = .396240625*(3-x) which becomes:
x-4 = .396240625*3 - .396240625*x
add 4 to both sides of this equation and add .396240625*x to both sides of this equation to get:
x + .396240625*x = .396240625*3 + 4 which becomes:
1.396240625*x = 5.188721876
divide both sides of this equation by 1.396240625 to get:
x = 3.716208927
plug that value for x in your original equation to get:
16^(3.716208927 - 4) = 3^(3 - 3.716208927) which becomes:
.455283067 = .455283067, confirming the value for x is good.
-----
7^(x - 2) = 5^x
divide both sides of this equation by 5^x to get:
7^(x-2)/5^x = 1
take log of both sides of this equation to get:
log((7^(x-2))/(5^x)) = log(1) which becomes:
log(7^(x-2)) - log(5^x) = log(1) which becomes:
(x-2)*log(7) - x*log(5) = log(1) which becomes:
x*log(7) - 2*log(7) - x*log(5) = log(1)
add 2*log(7) to both sides of this equation to get:
x*log(7) - x*log(5) = log(1) + 2*log(7)
factor out the x to get:
x*(log(7)-log(5)) = log(1) + 2*log(7)
divide both sides of this equation by (log(7)-log(5) to get:
x = (log(1)+2*log(7))/(log(7)-log(5)) to get:
x = 11.56654212
plug that value of x into your original equation to get:
7^(11.56654212 - 2) = 5^11.56654212 which becomes:
121525103 = 121525103, confirming the value for x is good.
-----
in order to solve these problems, you absolutely have to store intermediate results in memory that are not rounded and then use them in future calculations involving them. even if your calculator shows 8 to 10 significant digits, that's usually not enough to get an answer that is right on. storing the intermediate results in memory and then recalling those results as needed guarantees you'll get the closest accuracy that the calculator can give you.
-----
i think i solved every problem except for the second one which is:
5^(x - 1) = 3x
if this problem is:
5^(x - 1) = 3^x, then it can be solved the same way we solved the other ones.
as it stands, the x not in the exponent causes difficulties.
i'll try to solve as is to show you what i mean.
divide both sides of the equation by 3x to get:
5^(x-1)/3x = 1
take log of both sides of the equation to get:
log(5^(x-1)/3x) = log(1) which becomes:
log(5^(x-1)) - log(3x) = log(1) which becomes:
(x-1)*log(5) - (log(3) + log(x)) = log(1) which becomes:
x*log(5) - log(5) - log(3) - log(x) = log(1) which becomes:
x*log(5) - log(x) = log(1) + log(5) + log(3)
the problem is that you now have x outside the logs and inside the logs which can't be solved.
i suspect this problem is really:
5^(x-1) = 3^x
shown that way, it can be solved the same way we solved the other problems.
-----
here's a decent reference you can use to study up on logarithm operations. also use your text and go through some of the examples and exercises to get a feel for how it's done.
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut45_expeq.htm
here's some more references that can help you.
pick the ones that make the most sense to you.
http://www.sosmath.com/algebra/logs/log4/log4.html
http://tutorial.math.lamar.edu/Classes/Alg/SolveLogEqns.aspx
http://tutorial.math.lamar.edu/Classes/Alg/SolveExpEqns.aspx
http://www.algebra.com/algebra/homework/logarithm/L.lesson

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