SOLUTION: i dont know how to solve
(3^5x)(9^x^2)=27
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Question 70853: i dont know how to solve
(3^5x)(9^x^2)=27
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
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Interesting problem.
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First recognize that can be written as . By the rules of
exponents this can be expanded to and this multiplication can be
done by adding the exponents to get . We can now replace
by . So let's do that very thing. When the substitution is done, the
problem becomes:
.
.
Since these two terms have the common base of 3, the multiplication can be done by adding
the exponents to get:
.
.
Now let's take the log of both sides, and the equation becomes:
.
.
On the left side you can take the exponent away and make it the multiplier of the log term:
.
.
But on the right side the 27 can be replaced by and the equation becomes:
.
.
And now on the right side we can take the exponent out and use it as the multiplier of
the log to make the equation:
.
.
Note that the log(3) is a factor on both sides. If we divide both sides by log(3)
it cancels out and we are left with the "nice" equation:
.
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Subtract 3 from both sides:
.
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The left side factors to give:
.
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Note that if either of the two factors equals 0, the equation will be true. So we can
find the values of x by setting the factors equal to zero.
.
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Add 1 to both sides and then divide by 2 to find that is a solution.
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Now set the other factor equal to zero.
.
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Subtract 3 from both sides and you have:
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So 1/2 and -3 are the two values of x. If you plug them, one at a time, into the equation
you were given as the problem and work out the resulting equation using a calculator,
you will find that they both work.
.
Like I said at the start, "Interesting problem." Good practice too. I hope you can work
your way through all the steps.
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