took 2^(5x) as variable U, got U^3 + U^2 - U^4 = -128 stuck here, pls halp
B BroPo New member Joined Jul 17, 2020 Messages 2 Aug 19, 2020 #1 took 2^(5x) as variable U, got U^3 + U^2 - U^4 = -128 stuck here, pls halp
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 Aug 20, 2020 #2 The "rules of exponents"are 1) \(\displaystyle (a^x)(a^y)= a^{x+ y}\) and 2) \(\displaystyle (a^x)^y= a^{xy}\) If \(\displaystyle U= 2^{5x}\) then \(\displaystyle U^3= (2^{5x})^3= 2^{3(5x)}= 2^{15x}\). You are using the rule for \(\displaystyle 2^3U= (2^{5x})(2^3)\), not \(\displaystyle U^3\).
The "rules of exponents"are 1) \(\displaystyle (a^x)(a^y)= a^{x+ y}\) and 2) \(\displaystyle (a^x)^y= a^{xy}\) If \(\displaystyle U= 2^{5x}\) then \(\displaystyle U^3= (2^{5x})^3= 2^{3(5x)}= 2^{15x}\). You are using the rule for \(\displaystyle 2^3U= (2^{5x})(2^3)\), not \(\displaystyle U^3\).
Steven G Elite Member Joined Dec 30, 2014 Messages 14,561 Aug 20, 2020 #3 25x+3= 25x**23 = 8*25x=8U So you should arrive at 8U + 4U - 16U = -128 -4U = -128 U= 32 25x=32. Now you solve for 5x and then x.
25x+3= 25x**23 = 8*25x=8U So you should arrive at 8U + 4U - 16U = -128 -4U = -128 U= 32 25x=32. Now you solve for 5x and then x.
D Deleted member 4993 Guest Aug 20, 2020 #4 BroPo said: took 2^(5x) as variable U, got U^3 + U^2 - U^4 = -128 stuck here, pls halp View attachment 21106 Click to expand... Another way (sort of): \(\displaystyle 2^3 * 2^{5x} + 2^2 * 2^{5x} - 2^4 * 2^{5x} = -2^7\) \(\displaystyle 8 * 2^{5x} + 4 * 2^{5x} - 16 * 2^{5x} = -2^7\) \(\displaystyle -4 * 2^{5x} = -2^7\) \(\displaystyle 2^2 * 2^{5x} = 2^7\) \(\displaystyle 2^{5x} = 2^5\) x = ?
BroPo said: took 2^(5x) as variable U, got U^3 + U^2 - U^4 = -128 stuck here, pls halp View attachment 21106 Click to expand... Another way (sort of): \(\displaystyle 2^3 * 2^{5x} + 2^2 * 2^{5x} - 2^4 * 2^{5x} = -2^7\) \(\displaystyle 8 * 2^{5x} + 4 * 2^{5x} - 16 * 2^{5x} = -2^7\) \(\displaystyle -4 * 2^{5x} = -2^7\) \(\displaystyle 2^2 * 2^{5x} = 2^7\) \(\displaystyle 2^{5x} = 2^5\) x = ?