more 10 points being offered Part 2 please show your work so i can understand?

Question

solve each exponential equation. Express irrational solutions in exact form and as a decimal round to 3 decimal places:
1.0.3^(1+x)=1.7^((2x)-1)
2.3^(2x) +(3^x)-2 =0
3.(9^x) -3^(x+1) +1=0
4.2*49^x +11*7^x +5 =0

i tired some of it but i got lose so please show your work so i can know where i did that was wrong

Answer 1

# 1 is already done it for u
#2 3 4 i dunno sorrie

Answer 2

1.
0.3^(1+x)=1.7^((2x)-1) -- Given.
log.3[1.7^(2x-1)] = 1+x -- Def. of Logs.
(2x-1)log.3(1.7) = 1+x -- Prop. of Logs.
2xlog.3(1.7) + log.3(1.7) = 1+x -- Distributive property.
x[2log.3(1.7)+1] = 1-log.3(1.7) -- Distributive property
x = [1-log.3(1.7)] / [2log.3(1.7)+1] -- Division
x = [1 - log(1.7)/log(.3)] / [2log(1.7)/log(.3) + 1]
x = 12.154

2.
3^(2x) + 3^x - 2 =0 -- Given
Let 3^x = u. This will give us a quadratic
u^2 + u - 2 = 0
(u + 2)(u-1) = 0 -- Factor
u = -2 and 1 -- If ab = 0 then a and b = 0. Solve
-- First solution:
3^x = -2 -- Replace u with 3^x
log3(-2) = x -- Def. of logs
x = Unreal = 0.631 + 2.860i
-- Second Solution:
3^x = 1 -- Replace u with 3^x
log3(1) = x -- Def. of logs
x = 0 -- Properties of logs. logb(1) = 0

3.

9^x -3^(x+1) +1=0 -- Given
3^(2x) - (3^x)*3 + 1 = 0 -- Make into quadratic form
Let u = 3^x
u^2 - 3u + 1 = 0 -- Make into quadratic
u = [3 +/- sqrt(9 -4)]/2 -- Quad. Formula
u = [3 + sqrt(5)] and [3 - sqrt(5)]/2
-- First solution
[3 + sqrt(5)]/2 = 3^x
log3{[3+sqrt(5)]/2} = x
x = log{[3+sqrt(5)]/2}/log(3)
x = 0.876
-- Second solution
[3 - sqrt(5)]/2 = 3^x
log3{[3 - sqrt(5)]/2} = x
x = log{[3-sqrt(5)]/2}/log(3)
x = -0.876

4.
4.2*49^x +11*7^x +5 =0
4.2[7^(2x)] + 11(7^x) + 5 = 0
4.2u^2 + 11u + 5 = 0
u = [-11 +/- sqrt(121 - 104)]/8.4
u = [-11 +/- sqrt(37)]/8.4
7^x = [-11 +/- sqrt(37)]/8.4
x = log7{[-11 +/- sqrt(37)]/8.4}
x = log{[-11 +/- sqrt(37)]/8.4}/log(7)
-- First solution, +
x = Unreal = -0.275 + 1.614i
-- Second solution, -
x = Unreal = 0.365 + 1.614i

Answer 3

1. 0.3^(1+x)=1.7^((2x)-1)
log .0.3^(1+x) = log 1.7^((2x)-1)
(1+x) log .3 = (2x-1) log 1.7
1+xlog.3 = 2xlog 1.7 - log1.7
1 +log 1.7 = x(2log1.7 -log .3
(1+log1.7)/ (2log1.7 - log.3) = x

3^(2x) +(3^x)-2 =0
(3^x)^2 + 3^x - 2 =0
Let u = 3^x
u^2+u-2 = 0
(u+2)(u-1) = 0
u = 1 or -2
3^x = 1 --> x --> x = ln 1/ln 3 --> x = 0
3^x = -2 --> x = ln -2/ ln 3 which is rejected because -2 is not in domain of x
Solve remaining problems in similar fashion.
3.(9^x) -3^(x+1) +1=0
4.2*49^x +11*7^x +5 =0

Answer 4

1) Take logs
(1+x) ln(0.3) =((2x)-1) ln(1.7)
so
[ln(0.3) - 2ln(1.7) ] x = - [ ln(0.3) + ln(1.7) ]
x= - [ ln(0.3) + ln(1.7) ] / [ln(0.3) - 2ln(1.7) ]

2) 3) and 4) use a different method for 2) let Z = 3^x to get a quadratic in Z :-- Z^2 + Z - 2 =0

solve this as usual then find x = ln(Z) / ln(3)

Similarly for the others