SOLUTION: Solve: a) {{{10 ^(x/2) = (10 ^x) / (10 ^2)}}} b) {{{ 2 ^x + (2 ^(-x)) = 1}}} c) {{{ log(3x) = (log3)*(logx) }}} d) {{{ log(x^3) = (logx)^3}}}

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve: a) {{{10 ^(x/2) = (10 ^x) / (10 ^2)}}} b) {{{ 2 ^x + (2 ^(-x)) = 1}}} c) {{{ log(3x) = (log3)*(logx) }}} d) {{{ log(x^3) = (logx)^3}}}      Log On


   

Question 120724: Solve:
a) 10+%5E%28x%2F2%29+=+%2810+%5Ex%29+%2F+%2810+%5E2%29
b) +2+%5Ex+%2B+%282+%5E%28-x%29%29+=+1
c) +log%283x%29+=+%28log3%29%2A%28logx%29+
d) +log%28x%5E3%29+=+%28logx%29%5E3
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
a) {10 ^(x/2) = (10 ^x) / (10 ^2)}
10^(x/2) = 10^(x-2)
x/2 = x-2
x = 2x-4
x = 4
--------------
b) { 2 ^x + (2 ^(-x)) = 1}
2^x + 1/2^x = 1
Multiply thru by 2^x to get:
2^(2x) + 1 + 2^x
(2^x)^2 - 2^x+1 = 0
(2^x -1)(2^x-1) = 0
2^x = 1
x = 0
-------------
c) { log(3x) = (log3)*(logx) }
log(3) + log(x) = (log3)*(logx)
log(3) = log(x)*log(3)- log(x)
log(3) = logx [ log(3) -1]
logx = [log3]/[log(3) -1]
logx = -0.9125
x = 10^(-0.9125)
x = 0.1223
----------------------
d) { log(x^3) = (logx)^3}
3log(x) = (logx)^3
(logx)^3 - 3log(x) = 0
Factor:
log(x)[(logx)^2-3] = 0
log(x) = 0 or (logx)^2=3
x = 1 or logx = sqrt3 or logx=-sqrt3
x = 1 or x = 10^sqrt3 or x = 10^(-sqrt3)
x = 1 or x = 53.96 or x=0.0185
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Cheers,
Stan H.