Find the Solution of the Exponential Equation?How do I solve 5 x = 4 x + 1?I understand how to solve for x when there is one exponent, but I don't know how to solve when there is an exponent on both sides of the equation..

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Find the Solution of the Exponential Equation?How do I solve       5    x    =      4    x    +  1?I understand how to solve for   x when there is one exponent, but I don&#039;t know how to solve when there is an exponent on both sides of the equation..

jillrozellmzoej · Accepted Answer

If   a  &amp;gt;  1      a    x    =      e          ln      ⁡                        a          x                      =      e          x      ln      ⁡              a                ⟹              d              a        x                    d      x        =  ln  ⁡      a    ⋅      a    x    &amp;gt;  0Therefore       a    x   is strictly increasing in       R   and it follows it is injective.Now, Since       5    4    &amp;gt;  1, the exponential   (      5    4        )    x   is strictly increasing in       R  , and consequently so is  (      5    4        )    x    −  1Multiply by the strictly increasing       4    x   to get an injective function:  [  (      5    4        )    x    −  1  ]  ⋅      4    x    =      5    x    −      4    x  The conclusion is that there is at most one real   x that satisfies       5    x    −      4    x    =  1, which is equivalent to:      5    x    =      4    x    +  1This single   x is   1 as Tom Bombadil pointed out.

Find the Solution of the Exponential Equation? How do I solve 5 x </msup> =

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