Question 1103116
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(Note that the given equation, as you show it, is 5^x = 4^x + 1, or {{{5^x = 4^x+1}}}; that is much different than the equation you are really trying to solve, which is 5^x = 4^(x+1), or {{{5^x = 4^(x+1)}}}.  As another tutor on this site likes to say, "parentheses are free -- use them!")<br>
You have done all you can do with the logarithms; now you have an equation in just x which you can solve.<br>
Yes; distribute.   Then solve for x (hint gather all the terms with x on one side of the equation).<br>
Then note that the given answer, {{{log(4)/log(5/4)}}}, can be written, using laws of logarithms, as {{{log(4)/(log(5)-log(4))}}}.<br>
When you finish solving your equation for x, that is exactly what you should get.