Posted by **riya** on Monday, April 21, 2008 at 10:21am.

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 -
**bobpursley**, Monday, April 21, 2008 at 11:18am
all correct.

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