SOLUTION: I am given the equation 4^x+4=5^2x+5 and I have to answer it in the form of a log. My professor did not go over something like this so I have no idea on where to start the problem.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I am given the equation 4^x+4=5^2x+5 and I have to answer it in the form of a log. My professor did not go over something like this so I have no idea on where to start the problem.      Log On


   

Question 540310: I am given the equation 4^x+4=5^2x+5 and I have to answer it in the form of a log. My professor did not go over something like this so I have no idea on where to start the problem.
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
You are given to solve for x:
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4%5E%28x%2B4%29=5%5E%282x%2B5%29
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Any time you have an unknown in the exponent, one of the things you should think of is using logarithms. Logarithms have a very useful property for such problems. .
Let's take the logarithm of both sides. We'll use logarithms to the base 10, but we could use natural logs as well. Taking the logarithm of both sides results in:
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log%2810%2C4%5E%28x%2B4%29%29+=+log%2810%2C5%5E%282x%2B5%29%29
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A useful property of logarithms is that exponents can be brought out as the multiplier of the logarithm operator. Bringing these two logarithms out as multipliers results in the equation becoming:
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%28x+%2B+4%29%2Alog%2810%2C4%29+=+%282x+%2B+5%29%2Alog%2810%2C5%29
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Note that log%2810%2C4%29 is just a number that you can get from a scientific calculator. Enter 4 and press the "log" key. To six decimal places you should see the answer of 0.602056. [This is the exponent you raise the base 10 to that will give you 4.]
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Similarly log%2810%2C5%29 is also just a number. Use your calculator to finds its value. You should get (to six decimal places) 0.698970. [This is the exponent you raise the base 10 to that will give you 5.]
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Substitute these two values into the log equation to get:
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%28x+%2B+4%29%2A%280.602056%29+=+%282x+%2B+5%29%2A%280.698970%29
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From here on out, its a fairly straightforward algebra problem.
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Do the distributed multiplication on the left side by multiplying 0.602056 times each of the two terms in the parentheses to make the left side become as shown below:
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0.602056x+%2B+4%2A0.602056+=+%282x+%2B+5%29%2A%280.698970%29
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The 4 times 0.602056 multiplies out to 2.408224 and the equation simplifies to:
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0.602056x+%2B+2.408224+=+%282x+%2B+5%29%2A%280.698970%29
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Next, follow the same procedure of distributed multiplication on the right side. Multiply 0.698970 times each of the two terms in parentheses to get:
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0.602056x+%2B+2.408224+=+1.397940x+%2B+3.494850
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Move the constant from the left side by subtracting 2.408224 from both sides. (On the right side the 3.494850 has 2.408224 subtracted from it.) The equation then is simplified to:
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0.602056x+=+1.397940x+%2B+1.086626
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Then transfer the 1.397940x from the right side to the left side by subtracting 1.397940x from both sides. The equation becomes:
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-0.795884x+=+1.086626
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Solve for x by dividing both sides by -0.795884 to get:
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x+=+1.086626%2F-0.705584
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and this equals:
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x+=+-1.365607
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Let's try this in the original problem to see if it checks.
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Start with:
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4%5E%28x%2B4%29=5%5E%282x%2B5%29
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Substitute -1.365607 for x:
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4%5E%28-1.365607%2B4%29+=+5%5E%282%2A-1.365607%2B5%29
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Algebraically simplify the two exponents:
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4%5E%282.634393%29+=+5%5E%28-2.731214%2B5%29
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The exponent on the right side sums to give:
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4%5E%282.634393%29+=+5%5E%282.268786%29
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Use a scientific calculator with an x%5Ey function key to find the following simplifications to each side of this equation:
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38.553395=38.531273
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Pretty close to equal on both sides. The round off errors of the logarithms cause some differences. And I could have made some fat-fingered keying errors on the calculator also. Check my work. This near equality tells us that the value of x is accurate within a "reasonable" amount.
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[A little closer value of x (by carrying more decimal places on a calculator) is x = -1.36529380395]
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The process of bringing exponents outside of logarithms as multipliers is correct. Hope you can translate this discussion into an understanding of how you can handle exponents in which a variable appears.
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