# pre-calculus-check over work

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1). change to exponential form:
log2^256=8; 2^8=256
log0.0001=-4; 10^-4=0.0001

2). change to logarithmic form:
5^3=125; log5^125=3
4^-3=1/64; log4^1/64=-3

3). solve the equation:
3^(x-2)=27^(x+1)
3^(x-2)=(3^3)^(x+1)
3^(x-2)=3^(3x+3)
x-2=3x+3
-5=2x
-5/2=x
2^(x^(2)+1)=4^(x+2)
2^(x^(2)+1)=(2^2)^(x+2)
2^(x^(2)+1)=2^(2x+4)
x^(2)+1=2x+4
x^(2)-2x-3
(x+1)(x-3)
x=-1;x=3

• pre-calculus-check over work -

all correct.

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