1.) Find all \(x\) so that \($8^{x^2} \cdot 4^x = 2$\). Enter all the solutions, separated by commas.
2.)
Problem:
Find all real solutions to
\(2 \log_2 (x + 5) = \log_2 (x - 9) + \log_2 (x + 53) + 1 \)
Enter all the solutions, separated by commas.
+129896 8^(x^2) * 4^x = 2
[ 2^3 }^(x^2) * (2^2)^x = 2^1
2^(3x^2) * 2^(2x) = 2^1
2^( 3x^2 + 2x) = 2^1 the bases are the same......solve for the exponents
3x^2 + 2x = 1
3x^2 + 2x - 1 = 0
(3x - 1)(x +1) = 0
Setting both factors to 0 and solving for x produces
x = 1 / 3 or x = -1
+129896 Note : 1 = log2(2)
2log2 ( x + 5) = log2(x - 9) + log2(x + 53) + log2(2)
log2(x + 5)^2 = log2(x - 9) + log2(x + 53) + log2(2)
log2(x + 5)^2 = log2 [ (x - 9)(x + 53) * 2 ] ....so...
(x + 5)^2 = 2(x - 9)(x + 53)
x^2 + 10x + 25 = 2 ( x^2 + 44x - 477)
x^2 + 10x + 25 = 2x^2 + 88x - 954
x^2 + 78x - 979 = 0
(x - 11) ( x + 89) = 0
Setting each factor to 0 and solving for x produces
x = 11 (accept) or
x = -89 (reject as it doesn't give a real valued log )
