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5^(y+2)=1
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Why guess? IS this \[5^{y+2}=1\] or \[5^{y}+2=1\] In the first case - take logs of both sides. In the second case move the 'constant' terms to one side of the equation (careful with the signs) and THEN take logs (however - this will cause you a problem - so I guess the equation MUST be the first case)
the first one
take logs of both sides then use the following identitiy: \[\log(a)^{n}= nlog(a)\]
What's the question?
You just gave us an expression and not the question..
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it says solve and check
solving exponential equations
There is however a more direct solution \[n ^{x}=1\] implies x=0
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