Solving an exponential equation by logarithmization

An exponential equation is one in which the unknown variable or variables is/are in the exponent of a power. The exponential equations use basic knowledge of the exponential and logarithmic functions.

To solve them the following properties are used:

  • $$a^0=1$$ for any $$a$$.
  • Two potencies with the samepositivebasis and different from the unit are equal, if, and only if, its exponents are equal. That is to say: $$$2^a=2^b \Leftrightarrow a=b$$$
  • For any $$a \neq 0$$ and $$a\neq 1$$ we have: $$$a^x=b \Rightarrow x= \log_ab$$$

When it comes to solving an exponential equation it can have different forms and, because of that, there are different methods and transformations.

When the exponential equation to be solved is of the type $$a^{f(x)}=b$$ then it is possible to solve both sides by logarithmization if both members are positive. Namely, the properties of the logarithm are simply applied to find the value of $$f(x)$$.

$$$\displaystyle 2^{x+1}=6^\frac{3x}{2}$$$

Applying logartihm:

$$$\displaystyle \log_2(2^{x+1})=\log_2 \Big(6^\frac{3x}{2}\Big)$$$

Now, by means of the properties of the logarithm,

$$$\displaystyle \log_2(2^{x+1})=\displaystyle \log_2 \Big(6^{\frac{3x}{2}}\Big) \Rightarrow \log_2(2)^{x+1}=\log_2(6)^{\frac{3x}{2}} \Rightarrow (x+1)\log_22=\frac{3x}{2}\log_26$$$

$$$\Rightarrow (x+1)=\frac{3x}{2} \log_26$$$

We have turned the exponential equation into an equation of the first degree that we can solve. This is, by isolating $$x$$ we obtain:

$$$(x+1)=\frac{3x}{2}\log_26 \Rightarrow x-\frac{3 \cdot \log_2 6}{2}x=-1 \Rightarrow x\Big(1-\frac{3 \cdot \log_2 6}{2}\Big)=-1\Rightarrow$$$

$$$x=\frac{-2}{2-3\cdot \log_2 6}$$$

$$$\displaystyle 5^{2x-1}=7^{3-x} \Rightarrow 2x-1=\log_5(7^{3-x})=(3-x)\log_5 7 \Rightarrow x=\frac{3\log_5 7+1}{2+\log_5 7}$$$