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solve the equation
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\[e^2x-1=(e^3)^- x\] the x in the end is supposed to be with the negative signal but couldn't get it to work
its e^2x-1 = (e^3)^-x
im guessing the 2x - 1 is all part of the exponent otherwise this is not so easy to solve. anyway for the right hand side we can multiply exponents raised to an exponent (e^3)^-x = e^-3x => e^(2x-1) = e^-3x we can cancel the e by taking a log with base e (natural log) to both sides =>2x-1 = -3x add like terms and isolate x x = 1/5
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