Answer:
a) x ≈ -0.5958
b) x ≈ 1.1403
c) no solutions
Explanation:
You want to solve various exponential equations:
- 9^(3x-2) = 5^(7x -1)
- 2(3^(2x)) -5(3^x) = 7
- 3^x +1 +56(3^-x) = 0
Strategy
The first of these can be solve using logarithms to transform it to a linear equation. The other two can be solve by the substitution z=3^x. They each then become ordinary quadratic equations in z.
a) 9^(3x-2) = 5^(7x -1)
Taking logs, we have ...
(3x -2)log(9) = (7x -1)log(5)
x(3log(9) -7log(5)) = -log(5) +2log(9)
x = log(81/5)/log(729/78125) ≈ -1.2095/-2.0301
x ≈ -0.5958
b) 2(3^(2x)) -5(3^x) = 7
Using the substitution described above, this is ...
2z² -5z = 7
z² -(5/2)z = 7/2 . . . . . divide by 2
z² -(5/2)z +25/16 = (7/2) +(25/16) . . . . . . complete the square
(z -5/4)² = 81/16
z = 5/4 +9/4 . . . . . . z cannot be negative
x = log(3.5)/log(3)
x ≈ 1.1403
c) 3^x +1 +56(3^-x) = 0
z +1 +56/z = 0 . . . . . using z = 3^x
z² +z +56 = 0 . . . . . multiply by z
Descartes' rule of signs tells you this has no positive real solutions for z, so the original equation has no solutions. (The graph in the third attachment has no x-intercepts.)
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Additional comment
In general, equations consisting of a combination of polynomial and exponential expressions (as in (b) and (c)) cannot be solved algebraically. If the exponential expressions are related in a way that allows substitution (as here), then there may be algebraic methods that will work.
We like to write the equations in the form f(x) = 0, so a graphing calculator's display of the function zeros is a display of solutions to the equation. This is what we have done in the first three attachments.
Using the log expressions we developed, you can find the values of x to as many decimal places as your calculator provides. More significant digits are shown in the fourth attachment.
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