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solve for x: (e^4)^x*e^x^2=e^12
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\[(e ^{4})^{x}\times e ^{x ^{2}}=e ^{12}\]
did i make it clear now ?
you have e^(4x+x^2)=e^12 to be equal should x^2+4x=12 solve it and it is
\[e ^{4x+x ^{2}}=e ^{12}\] by exponent rules later just take logarithm of bouth sides and solve equation for x
oh thanks. i got it. can you help me with this aswel ? \[e ^{4x-9}=\frac{ 1 }{ e }\]
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here u have 4x-9=-1
\[1/e=e ^{-1}\] the rest is the same like in previous one
bec0use 1/e=e^-1
oh thank you guys :) appreciate it a lot
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