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My algebra textbook says that if the bases in an exponential equation are the same, then we can "cancel out" the bases and set the exponents equal to each other and then solve for $x$.

For example:

$2^{3x-8} = 2^4$

$3x-8 = 4$

$x = 4$

My question is twofold:

One, why are we allowed to "cancel out" the bases and set the exponents equal to each other when the bases are the same?

Two, why can't this method be done when the bases are NOT the same?

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We are allowed to cancel out the bases because exponential function is injective, indeed for any injective function $f(x)$ by definition

$$f(x_1)=f(x_2) \implies x_1=x_2$$

In the case of exponential function injectivity is guaranteed by the fact that exponental function $f(x)=a^x$ is strictly increasing for $a>1$ (or strictly decreasing for $0<a<1$).

If the base are different it is not the case since in general $a^x\neq b^x$ for $a\neq b$.

When basis are not the same we can convert one to proceed in the same way, that is for example

$$4^{f(x)}=7^{g(x)} \iff 4^{f(x)}=(4^{\log_4 7})^{g(x)}=4^{g(x)\log_4 7}$$

To give an example for which the cancellation doesn't work let consider

$$\sin (f(x))=\sin (g(x)) \not \Rightarrow f(x)=g(x)$$

indeed $\sin x$ function is not injective on its natural domain $\mathbb{R}$.

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  • $\begingroup$ Sorry, I don't comprehend the significance of exponential functions being injective. My current understanding of math is at a very basic algebra level, could you please explain further? $\endgroup$ – Slecker Oct 22 '18 at 16:25
  • $\begingroup$ @Slecker That's a foundamental concept. The fact is that the exponential function is one-to-one. That is at any value $y>0$ there exists one and only one value $x$ such that $a^x=y$. Refer also to the following LINK $\endgroup$ – user Oct 22 '18 at 16:29
  • $\begingroup$ @Slecker For such kind of function then $f(x_1)=f(x_2) \implies x_1=x_2$. $\endgroup$ – user Oct 22 '18 at 16:31
  • $\begingroup$ @Slecker When the function is continuous and stricktly monotonic injectivity is always achieved. Therefore are also injective $\log x \quad x>0$,$\arctan x$, $x^2 \quad x>0$ or $x<0$, $1/x \quad x>0$ or $x<0$ etc. $\endgroup$ – user Oct 22 '18 at 16:33
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If the bases are the same (and positive) then $a^x=a^y$ can be divided through by the non-zero number $a^y$ to obtain $a^{x-y}=1$ from which we can conclude that either $a=1$ (we assumed $a\gt 0$) or $x-y=0$.

[The function $a^x$ is monotonic for positive $a$ - it increases with increasing $x$ for $a\gt 1$, is constant for $a=1$ and decreases with increasing $x$ for $a\lt 1$ - this makes it injective for positive $a\neq 1$]

If the bases are not the same - well $4^2=16$ and $2^2=4$ are not equal, even though they have the same exponent. One can take logarithms to any convenient base, though, and in appropriate cases this will lead to a solution.

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  • $\begingroup$ Very nice explanation! $\endgroup$ – user Oct 22 '18 at 16:17
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Be careful with complex numbers, for example: $$ 2^i = 2^{i \cdot (1+2\cdot ln(2)\cdot\pi)} $$ As: $$ 2^0 = 1 = e^{2\cdot\pi \cdot i} = 2^{2\cdot ln(2)\cdot\pi \cdot i} $$ But: $$ 0 \neq 2\cdot ln(2)\cdot\pi \cdot i$$ And: $$ 2 \neq 2^{1+2\cdot ln(2)\cdot\pi} $$

The exponential function forms a bijection between $R$ and $R^+$.

But in general, the exponential (and the natural logarithm) is not a bijective function in $C$ to $C$.

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