I need help with 43 and 68 step by step 19 4 x 1 90 log20 91

I need help with #43 and #68 step by step.

19. 4) x 1) 90. log20 91. In(x 2) 92. In x 5) 24, 10\' Practice Plu 93 In 17 32. 4e 10,273 31 3v 1977 95. in xl 36, ear 7 11.243 2- 10,476 gg. In 2x 38, 137 100. In 3 0.2 101. 12 42. 102. 3 12 41. 5 43, e 3er 2- 0 3eze 18 45, e 5eza 24 0 48. 22u 2 12 0 103. The Cal Solve each logarithmic equation in Exercises 49-92. Be sure to reject any value of r thar is not in the domain of the original logarithmic expressions Give the exact answer Then, where necessary use a calculauor to obtain a decimal approximation, correct to two decimal places for the solution. 104. T 50. log5x 49, log 52, in x 51. In r 54 logs(x 7)- 2 53, loga (x 5) 3 56. log2(x 50) 5 The f 55, log2(x 25) 4 58. log 7(x 2) -2 sunli 57, log3(r 4) -3 of th 60. log2(4x 1 5 59, log (3r 2) 62, 6 ln(2x) 30 61. 5 ln (2x) 20 7 3 In x 63, 6 2 In x 65. Invr 3- 1 67, logs r logs(4x 1) 68. log6(r 5) log 6 x 2

Solution

43) Let e^x = t; thus the equation becomes:
t²-3t +2 =0;
this is a quadratic equation, thus we factorise it to find \'t\'
t² - t -2t+2=0;
t(t-1) -2(t-1) =0
(t-2)(t-1) =0 so
t=1 or t=2
if t =1 then e^x =1 meaning x= ln 1 =0; Thus x=0 (OR)
if t =2 then e^x=2 so x= log_e 2 = 0.6931 *from ln table)

Thus x=0 or x=0.6931

68) log₆(x+5) + log ₆ x =2;
we know that log (ab) = log a + log b; applying this formula we have;
log₆ (x+5) + log ₆x = log₆ (x*(x+5)) = log₆ (x²+ 5x) = 2 ;
This means that x² + 5x = 6² = 36
x² + 5x - 36 =0; this is a quadratic equation so solve for the roots by factorising;
x² + 9x-4x -36 =0\'
x(x+9) - 4 (x+9) =0
(x-4) ( x+9) =0 so
x=4 or x = 9 ;

Note: Once you have arrived at the formula of a quadratic equation, you can even apply the formula
x₁ = [-b + (b²-4ac) ] /2a &
x₂ = [-b - (b²-4ac) ] /2a

to get the roots of teh quadratic equation;