Section 6.6 Solutions

Section 6.6 solutions #1 - 12: Solve the exponential equation by writing each side of the equation with the same base then equating the exponents. Problems also may be solved with logarithms. 1) 2x = 16 (I will rewrite the 16 as 24) 2x = 24 Answer: x = 4 3) 2x+1 = 32 (rewrite the 32 as 25) 2x+1 = 25 x+1=5 Answer: x = 4 5) ( ) (easier to do with logs than my method) Rewrite (2-1)x = 24 2-x = 24 -x = 4 Answer: x = -4 7) (rewrite 64 as 26) 24-x = 26 4–x=6 -x = 2 Answer: x = -2 9) 32x = 2 (rewrite 32 as 25) (25)x = 21 25x = 21 5x = 1 Answer: x = 1/5 11) 16x = 4 (rewrite 16 as 42) (42)x = 41 42x = 41 2x = 1 Answer: x = 1/2 #13 - 24: Solve the exponential equations, round your answer to 2 decimals. 13) log (3x )= log 6 (take log of each side) x[log(3)] = log 6 ( use power to product rule) (divide then use calculator) Answer: x = 1.63 15) ex = 12 (take ln of each side) Ln ex = ln 12 x(ln e)= ln 12 (use power to product rule) x = ln 12 (ln e = 1) Answer: x = ln(12) = 2.48 17) 5(10x) = 20 (divide both sides by 5) 10x = 4 (now take log of each side) Log 10x = log 4 (use power to product rule) x(log 10) = log 4 (log 10 = 1) x = log 4 Answer: x = log(4) = .60 19) 32e2x = 128 (divide each side by 32) e2x = 4 (take ln of each side) ln e2x = ln 4 (use power to product rule) 2x(ln e)= ln 4 (ln e = 1) 2x = ln 4 (divide by 2) Answer: x = = .69 21) 3x-1 = 5x (take log of both sides) Log(3x-1) = log(5x) (use power to product rule) (x-1)log(3) = x(log 5) x( log 3) – log(3) = x( log 5) (move the xlog(5) to left and –log(3) to right) x(log 3) – x(log 5)= log(3) (factor out GCF of x) x( log 3 – log 5 )= log(3) (divide) Answer: 23) 6x-1 = 9x (take log of both sides) Log(6x-1) = log(9x) (use power to product rule) (x-1)log 6 = x(log 9 ) x(log 6) – log(6) = x(log 9) (move the x(log 9) to left and –log(6) to right) x(log 6) – x(log 9) = log(6) (factor out GCF of x) x( log(6) – log(9) ) = log(6) (divide) Answer: #25 - 51: Solve the logarithmic equations, round to 2 decimals when needed. 25) log3 x = 2 32 = x Answer: x = 9 27) ln x = 1 (write ln x as logex, then write in exponential form) Log ex = 1 Answer: x = e1 = 2.72 29) logx 49 = 2 (x >0) x2 = 49 (take the square root of both sides, don’t need since answer can only be positive) Answer: x = 7 31) logx 64 = 3 (x >0) x3 = 64 (now take cubed root of each side) √ √ Answer: x = 4 33) logx 3 = ⁄ (x>0) (write as square root since ½ power and square root are equivalent) √ (√ ) Answer: x = 9 35) log2(x-1) = 3 23 = x-1 8=x–1 Answer: x = 9 37) log2(2x)=5 25 = 2x 32 = 2x Answer: x = 16 39) log (x+1) = log (3x–2) (drop logs, set arguments equal) x + 1 = 3x – 2 -2x = -3 Answer: x = 3/2 41) log2 (x+3) = log2 (3x) (drop logs, set arguments equal) x + 3 = 3x 3 = 2x Answer: x = 3/2 43) log 2x – log2(x+6)=-2 (use minus to divide rule to rewrite left side with one log) (write in exponential form) (write 2-2 as ½2) (cross multiply) 4x = x + 6 3x = 6 Answer: x = 2 45) log2 x – log2 (x – 6) = 2 (use minus to divide rule to rewrite left side with one log) (write in exponential form) (rewrite the 22 as 4, then as a fraction over 1) (cross multiply) 4x – 24 = x 3x = 24 Answer: x = 8 47) log2 (x+6) – log2 (3x+2)=-1 (use minus to divide rule to rewrite left side with one log) (write 2-1 as ½1) (cross multiply) 3x + 2 = 2x + 12 Answer: x = 10 49) log3 x + log3 (x+6) = 3 (use the plus to multiply rule to rewrite left side with one log) Log3 (x)(x+6) = 3 Log3(x2 + 6x) = 3 (write in exponential form) 33 = x2 + 6x 27 = x2 + 6x 0 = x2 + 6x – 27 0 = (x+9)(x-3) x+9=0 x–3=0 x = -9 x=3 Check x = -9 Log3(-9) + log3(-9+6) =3 I get i on the left side, so this doesn’t check and -9 is not a solution Check x = 3 Log3 3 + log3 (3+6) = 3 1 + 2 = 3 (3 is a solution) Answer: x = 3 (no need to mention the x =-9 as it doesn’t check It would be okay to write: Answer: x = 3, x = -9 is extraneous) 51) log3 (x+6) + log3 (3x) = 4 (use the plus to multiply rule to rewrite left side with one log) Log3(x+6)(3x) = 4 Log3 (3x2 + 18x) = 4 34 = 3x2 + 18x 81 = 3x2 + 18x (divide every term by 3 to make numbers smaller) 27 = x2 + 6x 0 = x2 + 6x – 27 0 = (x+9)(x-3) x+9=0 x–3=0 x = -9 x=3 Check x = -9 Log3 (-9+6) + log3 (3*-9) = 4 Each of the logs gives an i, this solution doesn’t check. -9 is not a solution Check x = 3 Log3(3+6) + log3 (3*3) = 4 2 + 2 = 4, this solution check and x = 3 is the only solution to this problem. Answer: x = 3 (no need to mention the x =-9 as it doesn’t check It would be okay to write: Answer: x = 3, x = -9 is extraneous)