document.write( "Question 36187: how do you solve this problem : 2e and then in the exponent log base 3 (x +2) = 14? \n" ); document.write( "

Algebra.Com's Answer #26208 by astromathman(21) $\"\"$ $\"About$

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$\"2e%5E%283%28x%2B2%29%29+=+14\"$ . Take ln of both sides, remembering that $\"ln%28ab%29=ln%28a%29%2Bln%28b%29\"$ : $\"ln2%2Bln%28e%5E%283%28x%2B2%29%29%29=ln14\"$ , then, remembering that $\"ln%28x%29\"$ and $\"e%5Ex\"$ are inverse functions, $\"ln2%2B3%28x%2B2%29=ln14\"$ . From here just treat it as an ordinary Alg I equation and solve for x, keeping in mind that ln2 and ln14 are simply two numbers.
\n" ); document.write( "Solution:
\n" ); document.write( " $\"x=%28ln14-ln2%29%2F3-2\"$ \n" ); document.write( "

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