# How do you solve logarithmic equations like this one?

How do you solve $$3\log(x-15)=\left(\frac{1}{4}\right)^x?$$ The solution is approximately $16$. How would you solve a logarithmic equation with an solution approximately equal to a number without using a graphing calculator?

• Once upon a time, men didn't have cellphones but were smart. They knew a continuous function that changes sign should intersect the x-axis somewhere in between. They would notice that if you make $x=16$ the left hand side is zero while the right hand side stays above zero. They would then put $x=17$ and notice the left hand side is $3\log(2)=\log(8)>1$, while $1/4^{17}<1$. – OR. Dec 9 '13 at 1:11
• Men did not have cellphones? You are joking, right? – Igor Rivin Dec 9 '13 at 1:28
• No, it is true. (en.wikipedia.org/wiki/Mobile_phone#History) – OR. Dec 9 '13 at 1:35
• Only women had cell phones then. – dfeuer Dec 9 '13 at 1:39
• But really, it seems most unlikely to have a nice answer at all. Comparing a logarithm to an exponential will not generally turn out well, and the logarithm of a sum does not make it any prettier. – dfeuer Dec 9 '13 at 1:40

You don't say what base of logs you are using. The approach will be the same in any case-I will assume natural logs. We must have $x \gt 15$ or the logarithm is not defined. In that case, the right side will be very small and positive. We need $x \gt 16$ to make the left side positive. Define $a=x-16$, where we expect $a$ to be very small, so we will use the first term of the Taylor series. $$3 \log (x-15)=\left(\frac 14\right)^{\!x}\\ 3 \log (1+a)=\left(\frac 14\right)^{\!16+a}\\ 3\cdot 4^{16}a=4^{-a}$$ This shows $a \approx \frac 1{3\cdot 4^{16}}$ with both sides very close to $1$, so $x \approx 16+\frac 1{3\cdot 4^{16}}$.
The point is that for $x>10$ the right hand side is smaller than $1/1000000,$ so you are essentially solving for $LHS = 0.$ There is no general method if you are not so lucky.
• And the left hand side doesn't even exist for $x\le 15$ ... – hmakholm left over Monica Dec 9 '13 at 1:25