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Topos Uranos
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Exponential and logarithmic equations and inequalities

Summary

This lesson teaches how to solve equations and inequalities in which the unknown appears in an exponent or in the argument of a logarithm. It rests on two properties of the exponential and logarithmic functions: their injectivity, which allows exponents and arguments to be equated, and their monotonicity, which preserves or reverses the direction of an inequality according to the base. With them we study reduction to a common base, taking logarithms, the change of variable that turns an equation into a quadratic one, the domain of logarithmic equations and the extraneous solutions that appear when the laws of logarithms are applied outside their domain, systems, equations that no algebraic method solves, whose unique solution is located by bisection, and the analysis of the number of solutions of an equation with a parameter.

Learning objectives

  1. Solve exponential equations by reducing to a common base, by taking logarithms or by a change of variable that turns them into quadratic equations, discarding the nonpositive values of the new variable.
  2. Solve logarithmic equations by first determining their domain, applying the laws of logarithms only where they hold and recognizing extraneous solutions.
  3. Solve exponential and logarithmic inequalities, preserving or reversing the inequality according to the base, and express the solution set as a union of intervals.
  4. Prove, with the intermediate value theorem, that an equation without an algebraic method has a unique solution, locate it by bisection and analyze the number of solutions of an equation with a parameter.

Exponential equations

An exponential equation is an equation in which the unknown appears in some exponent, such as 2x=3x−12^x = 3^{x - 1} or 4x−2x−6=04^x - 2^x - 6 = 0. To solve them we use the facts studied in the lessons on the exponential function and on the logarithm function. If a>0a > 0 and a≠1a \neq 1, the power axa^x is defined for every real xx and is always positive; it is strictly increasing if a>1a > 1 and strictly decreasing if 0<a<10 < a < 1, and it takes each positive value exactly once. Its inverse is the logarithm to base aa, defined on the positive reals, which satisfies the product, quotient and power laws, provided the arguments are positive, and the change of base formula.

TheoremInjectivity of the exponential and of the logarithm

Let aa be a positive real different from 11. For all reals uu and vv, we have au=ava^u = a^v if and only if u=vu = v; for all positive reals uu and vv, we have log⁡au=log⁡av\log_a u = \log_a v if and only if u=vu = v. Moreover, if c>0c > 0, the equation ax=ca^x = c has exactly one solution, namely x=log⁡acx = \log_a c, and if c≤0c \leq 0 it has none.

We use the strict monotonicity of the exponential and the definition of the logarithm as its inverse. The figure follows the case a=2a = 2: the horizontal line at height 33 meets the graph of y=2xy = 2^x at a single point, and the one at height −2-2 does not meet it.

Proof

  1. u≠v→au≠av,au=av↔u=vu \neq v \rightarrow a^u \neq a^v, \qquad \dato{ie}{a^u = a^v \leftrightarrow u = v}

    If u=vu = v, the powers coincide, because a function assigns to each number a single value. Conversely, if u≠vu \neq v, one of the two is smaller, and by the strict monotonicity of the exponential, proved in the theorem on the properties of the exponential function, their powers are different.

  2. log⁡au=log⁡av→u=alog⁡au=alog⁡av=v\log_a u = \log_a v \rightarrow u = a^{\resaltar{\log_a u}} = a^{\resaltar{\log_a v}} = v

    If log⁡au=log⁡av\log_a u = \log_a v, we apply to both sides the exponential with base aa, which undoes the logarithm, and we get u=vu = v; the converse is evident.

  3. ax=c=alog⁡ac↔x=log⁡aca^x = c = a^{\resaltar{\log_a c}} \leftrightarrow x = \resaltar{\log_a c}

    If c>0c > 0, the number log⁡ac\log_a c is a solution, by the definition of the logarithm; and it is the only one, because every solution has the same power as it does.

  4. ax>0≥ca^x > 0 \geq c

    If c≤0c \leq 0, there is no solution, because the power axa^x is positive for every real xx.

Reducing to a common base

If both sides can be written as powers of a common base, injectivity allows the exponents to be equated, and the exponential equation becomes an algebraic equation. For example, since 8=238 = 2^3 and 4=224 = 2^2, the equation 8x−1=4x+18^{x - 1} = 4^{x + 1} is equivalent to 23x−3=22x+22^{3x - 3} = 2^{2x + 2}, that is, to 3x−3=2x+23x - 3 = 2x + 2, whose solution is x=5x = 5:

8x−1=4x+1↔3x−3=2x+2↔x=58^{x - 1} = 4^{x + 1} \leftrightarrow 3x - 3 = 2x + 2 \leftrightarrow x = 5

Taking logarithms

When the bases are not powers of a common number, such as 22 and 33, we take the logarithm of both sides. If both sides are positive, the injectivity of the logarithm guarantees that the resulting equation is equivalent to the original one: no solutions are lost or added. Any base will do; ordinarily one uses the natural logarithm, defined in the lesson on the logarithm function, or the logarithm to base 1010.

ExampleAn equation with two bases

Solve the equation 2x=3x−12^x = 3^{x - 1}.

Proof

  1. 2x=3x−1↔ln⁡(2x)=ln⁡(3x−1)2^x = 3^{x - 1} \leftrightarrow \resaltar{\ln(2^x)} = \resaltar{\ln(3^{x - 1})}

    Both sides are positive for every xx. We take the natural logarithm of both, which gives an equivalent equation by the injectivity of the logarithm.

  2. ln⁡(2x)=ln⁡(3x−1)↔xln⁡2=(x−1)ln⁡3\resaltar{\ln(2^x)} = \resaltar{\ln(3^{x - 1})} \leftrightarrow \dato{e}{x \ln 2 = (x - 1) \ln 3}

    The power law brings the exponents down as factors.

  3. xln⁡2−(x−1)ln⁡3=0↔x(ln⁡3−ln⁡2)=ln⁡3x \ln 2 - \resaltar{(x - 1) \ln 3} = 0 \leftrightarrow \resaltar{x(\ln 3 - \ln 2)} = \ln 3

    We gather the terms with xx on one side. The coefficient ln⁡3−ln⁡2\ln 3 - \ln 2 is positive, because the natural logarithm is strictly increasing and 2<32 < 3.

  4. x=ln⁡3ln⁡3−ln⁡2=log⁡323x = \frac{\ln 3}{\resaltar{\ln 3 - \ln 2}} = \dato{s}{\log_{\frac{3}{2}} 3}

    We divide by that coefficient. By the quotient law, the denominator is ln⁡32\ln \frac{3}{2}, and by the change of base the solution is a logarithm to base 32\frac{3}{2}; it is approximately 2.71.

  5. 22=4>31,23=8<322^2 = 4 > \resaltar{3^1}, \quad 2^3 = 8 < \resaltar{3^2}

    A table of values confirms the location: at 22 the first power is greater, and at 33 the second; the graphs cross between these two numbers, at the only point at which xx satisfies the equation.

Equations that reduce to a quadratic

If axa^x and a2xa^{2x} appear in the equation, or powers that reduce to them, such as 4x=(2x)24^x = (2^x)^2, the change of variable t=axt = a^x turns it into a quadratic equation in tt, which is solved with the methods of the lesson on factoring quadratic polynomials. Now, the new variable does not range over all the reals: since axa^x is always positive, only the positive values of tt are admitted. Each positive root gives, by the injectivity theorem, exactly one solution x=log⁡atx = \log_a t, and each negative or zero root is discarded.

ExampleA change of variable

Solve the equation 4x−2x−6=04^x - 2^x - 6 = 0.

Proof

  1. 4x=(22)x=(2x)2,t=2x>0\resaltar{4^x} = (2^2)^x = \resaltar{(2^x)^2}, \quad \dato{t}{t = 2^x > 0}

    We write 4x4^x as the square of 2x2^x and call tt this power, which is positive.

  2. t2−t−6=(t−3)(t+2)=0\resaltar{t^2 - t - 6} = \resaltar{(t - 3)(t + 2)} = 0

    The equation is quadratic in tt, and it is factored by looking for two numbers with product −6-6 and sum −1-1.

  3. t=−2→2x=−2<0t = -2 \rightarrow 2^x = \resaltar{-2} < 0

    The root t=−2t = -2 is discarded, because tt is positive: the equation 2x=−22^x = -2 has no solution. The figure shows the parabola in the plane of the variable tt, of which only the portion with t>0t > 0 counts.

  4. 2x=3↔x=log⁡232^x = \resaltar{3} \leftrightarrow x = \resaltar{\log_2 3}

    The root t=3t = 3 is positive and gives exactly one solution, by the injectivity theorem; it is approximately 1.58.

  5. 4log⁡23−2log⁡23−6=9−3−6=04^{\log_2 3} - 2^{\log_2 3} - 6 = \resaltar{9 - 3 - 6} = \resaltar{0}

    Check: by the definition of logarithm, 2log⁡23=32^{\log_2 3} = 3, and the square of that power is 99.