document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #353693 by bucky(2189)\"\" \"About 
You can put this solution on YOUR website!
You are given to solve for x:
\n" ); document.write( ".
\n" ); document.write( "\"4%5E%28x%2B4%29=5%5E%282x%2B5%29\"
\n" ); document.write( ".
\n" ); document.write( "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. .
\n" ); document.write( "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:
\n" ); document.write( ".
\n" ); document.write( "\"log%2810%2C4%5E%28x%2B4%29%29+=+log%2810%2C5%5E%282x%2B5%29%29\"
\n" ); document.write( ".
\n" ); document.write( "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:
\n" ); document.write( ".
\n" ); document.write( "\"%28x+%2B+4%29%2Alog%2810%2C4%29+=+%282x+%2B+5%29%2Alog%2810%2C5%29\"
\n" ); document.write( ".
\n" ); document.write( "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.]
\n" ); document.write( ".
\n" ); document.write( "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.]
\n" ); document.write( ".
\n" ); document.write( "Substitute these two values into the log equation to get:
\n" ); document.write( ".
\n" ); document.write( "\"%28x+%2B+4%29%2A%280.602056%29+=+%282x+%2B+5%29%2A%280.698970%29\"
\n" ); document.write( ".
\n" ); document.write( "From here on out, its a fairly straightforward algebra problem.
\n" ); document.write( ".
\n" ); document.write( "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:
\n" ); document.write( ".
\n" ); document.write( "\"0.602056x+%2B+4%2A0.602056+=+%282x+%2B+5%29%2A%280.698970%29\"
\n" ); document.write( ".
\n" ); document.write( "The 4 times 0.602056 multiplies out to 2.408224 and the equation simplifies to:
\n" ); document.write( ".
\n" ); document.write( "\"0.602056x+%2B+2.408224+=+%282x+%2B+5%29%2A%280.698970%29\"
\n" ); document.write( ".
\n" ); document.write( "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:
\n" ); document.write( ".
\n" ); document.write( "\"0.602056x+%2B+2.408224+=+1.397940x+%2B+3.494850\"
\n" ); document.write( ".
\n" ); document.write( "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:
\n" ); document.write( ".
\n" ); document.write( "\"0.602056x+=+1.397940x+%2B+1.086626\"
\n" ); document.write( ".
\n" ); document.write( "Then transfer the 1.397940x from the right side to the left side by subtracting 1.397940x from both sides. The equation becomes:
\n" ); document.write( ".
\n" ); document.write( "\"-0.795884x+=+1.086626\"
\n" ); document.write( ".
\n" ); document.write( "Solve for x by dividing both sides by -0.795884 to get:
\n" ); document.write( ".
\n" ); document.write( "\"x+=+1.086626%2F-0.705584\"
\n" ); document.write( ".
\n" ); document.write( "and this equals:
\n" ); document.write( ".
\n" ); document.write( "\"x+=+-1.365607\"
\n" ); document.write( ".
\n" ); document.write( "Let's try this in the original problem to see if it checks.
\n" ); document.write( ".
\n" ); document.write( "Start with:
\n" ); document.write( ".
\n" ); document.write( "\"4%5E%28x%2B4%29=5%5E%282x%2B5%29\"
\n" ); document.write( ".
\n" ); document.write( "Substitute -1.365607 for x:
\n" ); document.write( ".
\n" ); document.write( "\"4%5E%28-1.365607%2B4%29+=+5%5E%282%2A-1.365607%2B5%29\"
\n" ); document.write( ".
\n" ); document.write( "Algebraically simplify the two exponents:
\n" ); document.write( ".
\n" ); document.write( "\"4%5E%282.634393%29+=+5%5E%28-2.731214%2B5%29\"
\n" ); document.write( ".
\n" ); document.write( "The exponent on the right side sums to give:
\n" ); document.write( ".
\n" ); document.write( "\"4%5E%282.634393%29+=+5%5E%282.268786%29\"
\n" ); document.write( ".
\n" ); document.write( "Use a scientific calculator with an \"x%5Ey\" function key to find the following simplifications to each side of this equation:
\n" ); document.write( ".
\n" ); document.write( "\"38.553395=38.531273\"
\n" ); document.write( ".
\n" ); document.write( "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.
\n" ); document.write( ".
\n" ); document.write( "[A little closer value of x (by carrying more decimal places on a calculator) is x = -1.36529380395]
\n" ); document.write( ".
\n" ); document.write( "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.
\n" ); document.write( ".
\n" ); document.write( "
\n" ); document.write( "
\n" );