Solve the following exponential equation
2^{2X}\:\:-\:6\left(2^X\right)^{\:}\:+\:8\:=\:0
3^{2x}\:\:-\:4\left(3^{x+1}\right)\:+\:27\:=\:0
• 32x - 4(3x+1) + 27 = 0
Solve the following logarithmic equation
Log3\:\left(\:x^2\:+\:7x\:21\:\right)\:=\:2
Log8\:\left(x^2\:-\:8x\:+\:18\right)\:=\:1/3
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Expert's answer
2019-08-04T14:38:23-0400
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