Reviewing Algebra II - Part Two: Exponents & Solving Exponential Equations

  Simplifying and solving exponential equations is one of the more important aspects given the wide array of applications in the physical sciences, such as radioactive decay, obtaining the half life of an isotope etc. So in this Part Two we focus on reviving the skills involved in solving these equations - which are  fundamental part of Algebra II- or learning them for the first time. Again, this is also based on the rules and properties of manipulating exponents which are generally part of Algebra I.   For example, we already saw in the previous post one of these rules applied to quotients, i.e.

a ^m /a ⁿ     =  a ^{m - n}

^{So for:} 

a ⁵ /a ³   we get:    a ^{5 - 3}    =  a 

Other rules for exponents:

(i)  a ^-n  =  1/a ⁿ

(ii)   (a ^m) ⁿ     =  a ^m n 

(iii)   a ^1/n    =   ⁿ Ö a

(iv)   a ^m/ n     =   (a ^1/n ) ^m     = (a ^m) ^1/n

=  ( ⁿ Ö a ) ^m     =   ⁿ Ö( a ^m )

There are generally 2 ways to solve exponential equations: i) where both sides of the equation have the same base, and (ii) when they have different bases.

For case (i) consider:   

3 ^x =  3 ^{2x - 2}

^{Here the same base (3) allows us to solve for x by equating the exponents:}

^{x =  2x - 2} 

^{so:  x = 2   which can be checked by plugging x = 2 back into the eqn.  Then:}

^{3 2 =  3 2(2) - 2}

3 ² = 3²

When different bases are involved, we have basically 2 choices: a) work to get both sides to the same base, then proceed as shown for case (i) or b) crank out the needed value using logs, say base 10 or natural. Since most scientific work is now with the latter we use that:  For example, take the equation:  4^x = 15

We begin by taking the natural logarithm of both sides:

ln 4^x   =   ln 15   

Then use the power rule: i.e. ln b^x   = x ln b, to bring the variable exponent (x) to the front, i.e.

x ln 4 =  ln 15

We then can solve by dividing both sides by ln 4:

x =    ln 15/  ln 4  =   2.708/ 1.386  = 1.953

The other approach, usually easier, is to bring each side of an exponential equation to the same base. For example, say: 5^{x -1} = 0.04

The trick here is to get both sides in base 5 and go from there. It helps if we can see the decimal 0.04 can be expressed as a fraction, 0.04 = 4/ 100.

But 4/100 is just 1/ 25 and we immediately see  5² =  25, and we have our base 5.

Then,  working through we get:  5^{x -1} = 1/ 4^x = 15
 =    5^-2

So we can now equate the exponents:  x - 1 =  -2 or x = -2 + 1  = -1

Practice Problems: 

Solve each equation for the unknown and check:

 (1) 2a^-3 - 1 = 15

(2) 9^x =  55

3) 5x^1/2 + 7   = 22

4) 3a³   = 81

5) 4^x = 8 ^x+1

6) b^-5 = 1/32

 

7)  3 +   7 ^x-1  =  10

For each of the following, write as a power with positive exponents in simplest form:

8)

9)

10)