How would you solve for $x$ in $7^{x}=5^{x-4}$?

Question

How would you be able to make these the same base? I tried to put it into a base $10$ formula, but it ended up making $x=0$.

Answer 1

Since we only have linear powers of $x$ (that is, powers of the form $ax+b,$ where $a,b$ constants with $a\ne 0$), then one approach we can take (that puts off logarithms until the very end) is this: $$7^x=5^{x-4}\\7^x=\frac{5^x}{5^4}\\7^x=\frac{5^x}{625}\\7^x\cdot625=5^x\\625=\frac{5^x}{7^x}\\625=\left(\frac57\right)^x$$ At this point, we can take a logarithm (of any base we like) on both sides, and use power rule to isolate $x$. More simply, though, we may use the definition of a logarithm. That last equation can be thought of as saying "$x$ is the exponent on base $\frac57$ that yields $625.$" But this means precisely that $$x=\log_{\frac57}625.$$ If non-standard bases for logarithms are undesirable for you, then we can apply the change-of-base formula to see that $$x=\cfrac{\log625}{\log\frac57}$$ or $$x=\cfrac{\ln625}{\ln\frac57},$$ whichever you prefer. (We can also get by power rule as discussed earlier.)

---

Added: We can do the same kind of thing more generally, too. Suppose that $s,t$ are positive real numbers, $a,c$ non-zero real numbers, and $b,d$ any real numbers. Then the following are equivalent: $$s^{ax+b}=t^{cx+d}\\s^{ax}s^b=t^{cx}t^d\\(s^a)^xs^b=(t^c)^xt^d\\\frac{(s^a)^x}{(t^c)^x}=\frac{t^d}{s^b}\\\left(\frac{s^a}{t^c}\right)^x=\frac{t^d}{s^b}$$ At that point, we can use power rule of logarithms or the definition, as described above.

Answer 2

Take log on both sides (does not matter if you use natural or common logs)

$$ x \log(7) = (x-4) \log(5) $$

Solving for $x$

$$ x = \frac{-4 \log(5)}{\log(7)-\log(5)} \approx -19.133 $$

Why do I feel like I am doing someone's homework problem!

Answer 3

Let's try to transform the exponential expression on the left hand side (which has a base of 7) into an expression with the same base as the exponential expression on the right hand side (its base is 5).

Remember, a logarithm is an exponent. To complete the first part of your task, ask yourself: what exponent do I raise 5 by to get 7? By the definition of logarithm, this number is $\log_5 (7)$. Therefore, since $5^{\log_5 (7)}=7$, we have

$7^x=5^{x-4}\\ ( 5^{\log_5 (7)} )^x=5^{x-4} $

Now, remember what happens when you raise an exponential expression of a number by an exponent? You multiply the exponents! $(b^n)^m=b^{m*n}.$ Therefore,

$( 5^{\log_5 (7)} )^x=5^{x-4}$

now can be expressed as the equation

$5^{x \log_5 (7)} =5^{x-4}$

What can you say about an equation involving only exponentials with the same positive number as base on both sides of the equal sign?