Exponential and Logarithmic Functions (without)?

Question

Solve the equation without using logarithms.

1. 4^5x = 16^2x-1


Solve the equation, First express your answer in terms of natural logarithms (For instance, x=(2+ 1n 5) / (1n 3)). Then use a calculator to find an approximation for the answer.

2. 2^x = 3^x-1


3. e^2x = 5

Please show all work and formula used to get this answer.

I Thank You All in advance.

Answer 1

I will use Newton-Raphson method:
x_next = x - f(x)/f'(x)

Answer 2:
x_next = [2^x - 3^(x-1)] / [ln(2) * 2^x - ln(3) * 3^(x-1)]

Starting with x = 2 and iterating ten times, we get the following results
2
3.9111391257032
3.33790302298842
2.93723688117754
2.74820986846139
2.71080872895182
2.7095127977045
2.70951129135349
2.70951129135146
2.70951129135145

Answer 3:
x_next = x - (e^(2x) - 5)/(2e^(2x))

Starting with x = 0 and iterating ten times, we get the following results
2
1.54578909722184
1.15936459749437
0.905360976385077
0.814201048596556
0.804808300620266
0.804718964198997
0.80471895621705
0.80471895621705

Answer 2

I will do number 1 only.

4^5x = 16^(2x - 1)

The goal here is to get the same base on both sides of the equation and then solve for x by bringing down the exponents.

How can I write 16 using 4 as a base?

Well, 4^2 is the same thing as saying 16, right? This is because 4^2 = 4 x 4 = 16.

The left side says the same but the right side becomes
4^2(2x - 1)

We now have this equation:

4^5x = 4^2(2x - 1)....Do you see that we now have the SAME BASE 4 on both sides of the equation?

Bring down both exponents and equate them to solve for x.

5x = 2(2x - 1)

5x = 4x - 2

5x - 4x = -2

x = -2

Is this clear?

Answer 3

16^(2x-1) = 4^2(2x-1)

5x = 2(2x-1) = 4x -2
x = -2

Answer 4

1. 4^(5x) = 16^(2x-1)
2^(10x) = 2 ^(4(2x-1))
10x =8x-4
2x = -4
x = -2

2. 2^x = 3^(x-1)
ln (2^x) = ln(3^(x-1))
xln(2) = (x-1) ln(3) = xln(3) -ln(3)
xln(2) - xln(3) = - ln(3)
x(ln(2) -ln(3)) = -ln (3)
x = ln(3)/(ln(3)-ln(2) = approox 2.7095

3. e^2x = 5
2x = ln(5)
x = ln(5)/2 = approx .0847