Solving Exponential Equations

The unknown is an exponent

An exponential equation has the unknown in an exponent, as in $2^x=32$2ˣ = 32. The usual moves do not reach it. Dividing both sides by 2 gives $2^x/2=16$2ˣ / 2 = 16, which is $2^{x-1}=16$2ˣ⁻¹ = 16, and x is still in the exponent. Adding, subtracting, multiplying and dividing all act on the value of $2^x$2ˣ, not on the x inside it.

There are two ways to reach it. If both sides can be written as powers of the same base, compare the exponents. If they cannot, take logarithms, since a logarithm undoes raising to a power.

Same base: equate the powers

In $2^x=32$2ˣ = 32, the right side is a power of 2: $32=2^5$32 = 2⁵. So the equation says $2^x=2^5$2ˣ = 2⁵. The graph of $y=2^x$y = 2ˣ rises all the way from left to right, so it reaches each height only once: two powers of 2 are equal only when their exponents are equal. So x = 5.

Often a number has to be rewritten first. To solve $2^{x+1}=8$2ˣ⁺¹ = 8, write 8 as $2^3$2³: then $2^{x+1}=2^3$2ˣ⁺¹ = 2³, so x + 1 = 3 and x = 2. Check: $2^3=8$2³ = 8.

Sometimes both sides need rewriting. In $9^x=27$9ˣ = 27, 27 is not a whole power of 9, but both are powers of 3: $9=3^2$9 = 3² and $27=3^3$27 = 3³. The power of a power multiplies the exponents, so $9^x={\left(3^2\right)}^x=3^{2x}$9ˣ = (3²)ˣ = 3²ˣ, and the equation becomes $3^{2x}=3^3$3²ˣ = 3³. So 2x = 3 and $x=\frac{3}{2}$x = 3/2. Check: $9^{\frac{3}{2}}$9^(3/2) is the square root of 9, cubed, which is $3^3=27$3³ = 27.

The same works with the unknown on both sides. In $4^x=8^{x-1}$4ˣ = 8ˣ⁻¹, write $4=2^2$4 = 2² and $8=2^3$8 = 2³: $2^{2x}=2^{3\left(x-1\right)}$2²ˣ = 2^(3(x − 1)), so 2x = 3x − 3 and x = 3. Check: $4^3=64$4³ = 64 and $8^2=64$8² = 64.

The gold curve is $y=2^x$y = 2ˣ and the white line is the level 32. The curve rises all the way, so it meets the line only once, at (5, 32): $2^x=32$2ˣ = 32 has the one solution x = 5.

Different bases: take logs of both sides

Most equations cannot be written with one base. 40 is not a power of 2, and 20 is not a power of 3. For these, take the logarithm of both sides.

This is allowed for the same reason as adding 5 to both sides. The two sides of $2^x=32$2ˣ = 32 are the same number, so their logarithms are the same number too. The only condition is that the number is positive, since only positive numbers have logarithms, and a power of a positive base is always positive.

Take the common logarithm, to base 10, of both sides: $\log \left(2^x\right)=\log 32$log(2ˣ) = log 32. The power law of logarithms, $\log \left(a^n\right)=n\log a$log(aⁿ) = n log a, brings the exponent down to the front: x log 2 = log 32. Now x is an ordinary unknown multiplied by a number, log 2.

Then it is an ordinary division

Divide both sides by log 2: $x=\log 32/\log 2$x = log 32 / log 2. A calculator gives log 32 = 1.50515 and log 2 = 0.30103, to 5 decimal places, and 1.50515 ÷ 0.30103 = 5. The answer is exactly 5, because $\log 32=\log \left(2^5\right)=5\log 2$log 32 = log(2⁵) = 5 log 2. Check: $2^5=32$2⁵ = 32.

Here the logs only confirmed what the same-base method found, and $\log 32/\log 2$log 32 / log 2 is ${\log }_{2}32$log₂ 32 by the change of base rule. The method matters when no whole power works.

When no whole power works

Solve $2^x=40$2ˣ = 40. Since $2^5=32$2⁵ = 32 and $2^6=64$2⁶ = 64, x lies between 5 and 6, and no whole number of 2s multiplied together makes 40. The same two steps still work. Take logs of both sides: x log 2 = log 40. Divide by log 2: $x=\log 40/\log 2=1.60206/0.30103=5.32$x = log 40 / log 2 = 1.60206 / 0.30103 = 5.32, to 2 decimal places.

Check by putting it back: $2^{5.32}=39.95$2^5.32 = 39.95, to 2 decimal places, which is 40 to the accuracy of the rounded answer. With more decimal places, $2^{5.3219}=40.00$2^5.3219 = 40.00.

Any base of logarithm works, as long as both sides use the same one. With natural logarithms, $x=\ln 40/\ln 2=3.6889/0.6931=5.32$x = ln 40 / ln 2 = 3.6889 / 0.6931 = 5.32 again.

The gold curve is $y=2^x$y = 2ˣ and the white line is the level 40. The curve passes (5, 32) below the line and (6, 64) above it, and meets the line between them, at x = 5.32.

The unknown on both sides

Solve $3^x=2^{x+1}$3ˣ = 2ˣ⁺¹. The bases are different, so take logs of both sides: x log 3 = (x + 1) log 2. The whole exponent x + 1 comes down, so log 2 multiplies both of its terms: x log 3 = x log 2 + log 2.

Collect the x terms on one side: x log 3 − x log 2 = log 2, so x(log 3 − log 2) = log 2. Then $x=\log 2/\left(\log 3-\log 2\right)=0.30103/0.17609=1.71$x = log 2 / (log 3 − log 2) = 0.30103 / 0.17609 = 1.71, to 2 decimal places.

Check: $3^{1.71}$3^1.71 and $2^{2.71}$2^2.71 are both 6.54, to 2 decimal places.

A number in front of the power

Solve $5\times 3^x=200$5 × 3ˣ = 200. The 5 multiplies the power; it is not part of the base. Divide both sides by 5 first: $3^x=40$3ˣ = 40. Then take logs: $x=\log 40/\log 3=1.60206/0.47712=3.36$x = log 40 / log 3 = 1.60206 / 0.47712 = 3.36, to 2 decimal places.

Taking logs straight away also works, using the product law: log 5 + x log 3 = log 200, so $x=\left(\log 200-\log 5\right)/\log 3$x = (log 200 − log 5) / log 3, and log 200 − log 5 = log 40. What does not work is treating $5\times 3^x$5 × 3ˣ as $15^x$15ˣ: the power applies to the 3 alone.

Inequalities: watch the sign of the log

Questions that ask "when does it first drop below" lead to an inequality. Find the smallest whole number x with $0.8^x<0.1$0.8ˣ < 0.1.

Take logs of both sides. The logarithm keeps the order of positive numbers, because a bigger number needs a bigger power of 10, so the inequality stays the same way round: x log 0.8 < log 0.1.

Now divide by log 0.8. The number 0.8 is less than 1, so its logarithm is negative: log 0.8 = −0.09691, to 5 decimal places. Dividing both sides of an inequality by a negative number reverses it: $x>\log 0.1/\log 0.8=\left(-1\right)/\left(-0.09691\right)=10.32$x > log 0.1 / log 0.8 = (−1) / (−0.09691) = 10.32, to 2 decimal places.

So the smallest whole number is x = 11. Check the whole numbers on either side: $0.8^{10}=0.107$0.8¹⁰ = 0.107, which is not below 0.1, and $0.8^{11}=0.086$0.8¹¹ = 0.086, which is.

A quadratic in disguise

Some equations have two powers of the same base, as in $4^x-5\times 2^x+4=0$4ˣ − 5 × 2ˣ + 4 = 0. The key is that $4^x={\left(2^2\right)}^x={\left(2^x\right)}^2$4ˣ = (2²)ˣ = (2ˣ)². So if u stands for $2^x$2ˣ, the equation is $u^2-5u+4=0$u² − 5u + 4 = 0, an ordinary quadratic.

It factors as (u − 1)(u − 4) = 0, so u = 1 or u = 4. Then $2^x=1$2ˣ = 1 gives x = 0, and $2^x=4$2ˣ = 4 gives x = 2. Check: $4^0-5\times 2^0+4=1-5+4=0$4⁰ − 5 × 2⁰ + 4 = 1 − 5 + 4 = 0, and $4^2-5\times 2^2+4=16-20+4=0$4² − 5 × 2² + 4 = 16 − 20 + 4 = 0.

A root of the quadratic can be impossible. $4^x-2^x-6=0$4ˣ − 2ˣ − 6 = 0 becomes $u^2-u-6=0$u² − u − 6 = 0, which factors as (u − 3)(u + 2) = 0, so u = 3 or u = −2. But u is $2^x$2ˣ, and every power of 2 is positive, so $2^x=-2$2ˣ = −2 has no solution and u = −2 is thrown away. That leaves $2^x=3$2ˣ = 3, so $x=\log 3/\log 2=1.58$x = log 3 / log 2 = 1.58, to 2 decimal places. Check: $2^x=3$2ˣ = 3 makes $4^x=3^2=9$4ˣ = 3² = 9, and 9 − 3 − 6 = 0.

The usual mistakes

Dividing the numbers instead of their logs. $2^x=40$2ˣ = 40 does not give $x=40/2=20$x = 40 / 2 = 20. That treats $2^x$2ˣ as 2 × x, and $2^{20}$2²⁰ is over a million.

Giving the number instead of the power. $3^x=27$3ˣ = 27 asks which power of 3 makes 27, so x = 3, not 27.

Subtracting the logs. $x=\log 40/\log 2$x = log 40 / log 2, which is 5.32. log 40 − log 2 is $\log \left(40/2\right)=\log 20=1.30$log(40 / 2) = log 20 = 1.30, a different number.

Bringing down only part of the exponent. The log of $2^{x+1}$2ˣ⁺¹ is (x + 1) log 2, not x + log 2: log 2 multiplies both terms of the exponent.

Keeping the inequality the same way round after dividing by the log of a number below 1. That log is negative, so the sign reverses.