Question 127650
 7^(2x+3) = 5^(x-4)
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Use natural logs to solve this:
{{{ln(7^(2x+3)) = ln(5^(x-4))}}}
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Using the log equivalent of exponents, we can write it:
{{{(2x+3)*ln(7) = (x-4)*ln(5)}}}
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Find nat log (ln) of 7 and 5:
1.9459(2x+3) = 1.6094(x-4)
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Multiply what's inside the brackets
3.8918x + 5.8377 = 1.6094x - 6.4376
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Some basic algebra
3.8918x - 1.6094x = -6.4376 - 5.8377
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2.2824x = -12.2753
x = {{{(-12.2753)/2.2824}}}
x = -5.378 
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Check solution on a good calc:
enter 7^(2(-5.378)+3) = 2.7888
enter 5^(-5.378-4) = 2.7865, close enough to check our solution
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Did this make sense? You can use common logs (base 10) if you prefer, the method is the same. (the numbers will be different, but the answer will be the same)