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amishra
Solve for x: 3^x - 2 = 8/3^x
\[3^{x} - 2 = 8/3^{x}\]
multiply both sides by 3 to the x to eliminate the denominator and subtract eight from both sides\[3^{2x}-2*3^{x}-8=0\]
Yes, I got \[3^{x} = 4 , 3^{x} = -2\]
Now treat it as a quadratic equation set equal to zero and use the quadratic formula to solve for 3 to the x, because 3 to the x was squared, just like a variable, and we have second and third terms as well.\[\frac{ 2 +\sqrt{4-4*(-8)} }{ 2 }\] You can ignore the possibility where you subtract the square root, since that would give a negative answer, which 3 to the x can't equal. Solve and you get \[\frac{ 8 }{ 2 }=4=3^{x}\]Now you should take the natural logarithm of both sides and solve.\[\ln 4=\ln 3^{x}\]Pull the exponent down and divide both sides by the natural log of 3\[\ln 4=x \ln 3\]\[\frac{ \ln4 }{ \ln3 }=x\] That should be your answer
Thank you soo much!! That was very helpful! :D