how to solve difficult logarithmic equations

Find $ x $ if $ \log_2(3x+1) = 4 $. By the definition of the logarithm, \[\begin{align} 3x+1 &= 2^4 \\&= 16 \\ 3x &= 15 \\ x &= 5.\ _\square \end{align}\]

Another way to view this solution is that we took $ 2 $ to the power of the left side and got $ 3x+1 $, and took $ 2 $ to the power of the right side and got $ 2^4 = 16 $. More complicated logarithmic equations are often simplified by exponentiating both sides.

If $\log_{8}m + \log_{8}\frac{1}{6}=\frac{2}{3}$, then what is $m?$

Other equations can be simplified using other properties of logarithms. One difficulty that arises is that eliminating logarithms and solving the resulting equation can introduce spurious solutions. These solutions violate the principle that the argument of the log function must always be positive. It is generally wise to check solutions by plugging them into the original equation and making sure that both sides are defined.

Find all $ x$ such that $ \log_3(x-12) +\log_3(x-6) = 3. $ Simplify the given equation as follows: \[\begin{align} \log_3(x-12) + \log_3(x-6) &=3 \\ \log_3\big((x-12)(x-6)\big) &= 3\\ (x-12)(x-6) &= 27 \\ x^2-18x-45 &=0\\ (x-15)(x-3) &= 0. \end{align}\] This produces two potential solutions $ x=3, x=15$. But note that $ x = 3 $ is not an actual solution, as $ \log_3(3-12) $ is undefined. The only actual solution is $ x=15 $. $_\square$

The step in the solution above that was not reversible was the first one: although $ (x-12)(x-6)$ is positive if $ x =3$, $ x-12 $ and $ x-6$ are not themselves positive.

Find all real solutions $x$ to

\[3\log_2(x) - 1 = \log_2\left(\frac32 x-1\right).\]

Enter your answer as the sum of all such $ x $.

More complicated logarithmic equations often involve more than one base. It can help to introduce unknowns to solve for the logarithms first. Another useful identity is $ \log_x(y) = \frac{\log_z(y)}{\log_z(x)} $, especially since $ z$ can be chosen to be whatever simplifies the problem.

Suppose $ a,b$ are positive real numbers such that \[\log_a(10)+\log_b(100) = \log_{ab}(1000000).\] Show that $ a = b $ or $ a^2=b $. Rewrite this as \[\frac{\log(10)}{\log(a)} + \frac{\log(100)}{\log(b)} = \frac{\log(1000000)}{\log(ab)},\] where the logs are all to the base $ 10 $. This simplifies to \[\frac1{\log(a)}+\frac2{\log(b)} = \frac6{\log(ab)}.\] Let $ m = \log(a) $ and $ n = \log(b) $. Then $\log(ab) = m+n$, so \[\begin{align} \frac1{m}+\frac2{n} &= \frac6{m+n} \\ n+2m &= \frac{6mn}{m+n} \\ (2m+n)(m+n) &= 6mn \\ 2m^2+3mn+n^2 &= 6mn \\ 2m^2-3mn+n^2 &= 0 \\ (2m-n)(m-n) &= 0. \end{align}\] So $ m=n$ or $ 2m=n$. In the first case, $ \log(a)=\log(b) $, so $ a=b$. In the second case, $ 2\log(a) = \log(b) $, so $ \log(a^2)=\log(b) $, so $a^2=b $. $_\square $

The sum of all (positive) solutions of the equation

\[\log_{16}x +\log _x 16=\log_{512} x + \log_x {512}\]

can be written as $ \frac{a}{b} $, where $a$ and $b$ are coprime positive integers. What are the last three digits of $a+b$?

Equations involving exponents can often be simplified by taking logarithms:

Solve for $ x $ if \[\left( \frac{x}2 \right)^{\log_2(x)} = 8x.\] Take $\log_2$ of both sides: \[\begin{align} \log_2\left(\left(\frac{x}2 \right)^{\log_2(x)}\right) &= \log_2(8x) \\ \log_2(x)\log_2\left(\frac x2\right) &= \log_2(x)+3 \\ \log_2(x)\big(\log_2(x)-1\big) &= \log_2(x)+3, \end{align}\] and substituting $ y = \log_2(x) $ turns this into $ y(y-1)=y+3$, or $ y^2-2y-3 = 0 $. So $ y = 3 $ or $ y = -1 $, which leads to $ x = 8 $ or $ x = \frac 12 $. $_\square$

What is the sum of all possible real values of $x$ that satisfies the equation $x^{\log_{5} x} = \frac{x^3}{25}?$

Logarithms can also make computation easier in certain practical situations. For example, logarithms with base $10$ give information about the number of decimal digits in a number.

Given that \[\begin{align} \log_{10}(2) &= 0.3010299\ldots \\ \log_{10}(3) &= 0.4771212\ldots \\ \log_{10}(7) &= 0.8450980\ldots, \end{align}\] which is bigger, $ 12^{50} $ or $ 7^{64} $? How many decimal digits do these two numbers have? Compute \[\begin{align} \log_{10}\big(2^{100}3^{50}\big) &= 100\log_{10}(2)+50\log_{10}(3) \\ &= 30.10299\ldots + 23.85656\ldots \\ &= 53.9595\ldots \\\\ \log_{10}\big(7^{64}\big) &= 64\log_{10}(7) \\ &= 64(0.8450980\ldots) \\ &= 54.0862\ldots. \end{align}\] So $ 7^{64} $ is bigger. The number of decimal digits of $ x $ is $ \lfloor \log_{10}(x) \rfloor +1 $, so $ 2^{100}3^{50} $ has $ 54 $ digits and $ 7^{64} $ has $ 55 $ digits. $_\square$

Note that this is much easier than multiplying the numbers and comparing them directly. For applications that require large numbers (such as RSA encryption ), this is very important.

Find the number of digits in $\Large 9^{9^{2}}$.

Note : You may use the fact that $\log_{10} 3 = 0.4771$ correct up to 4 decimal places.

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Solving challenging logarithm equation

I came across this logarithm equation and can't seem to figure out how to solve it.

$$lg(2x-24)=2+ \frac 13 lg8 - \frac 14 x lg16$$

I only managed to simplify all the way till $$100=(x-12)(2^x)$$ (based on assumption of base 10)

Would gladly appreciate if anyone could suggest a trick to solve this.

Edit: Could give some benefit of doubt to whether the log is base 2 or 10. Any solutions are welcomed.

problem-solving

3 Answers 3

Assuming your logs are base $2$ we have $$\lg(2x-24)=2+ \frac 13 \lg8 - \frac 14 x \lg16\\ \lg(x-12)+1=2+1-x\\ lg(x-12)=2-x\\x-12=2^{2-x}\\ 2^x(x-12)=4$$ Clearly $x$ has to be just barely greater than $12$, so let $x=12+y$ Then we have $$y=\frac 1{1024\cdot 2^y}$$ This needs a numeric solution and is in a good form as the right side will change slowly with $y$. Let $y_0=0$ and iterate. After two iterations we have converged to $0.000975902$. Alpha will give you a solution of $y=\frac {W\left(\frac {\log 2}{1024}\right)}{\log 2}$in terms of the Lambert W function where these logs are natural logs.

Added: for base $10$ logs we can do the same. Again $x$ has to be a little greater than $12$ so write $x=12+y$ The equation becomes $$y=\frac {25}{1024\cdot 2^y}$$ which converges to $y\approx 0.0240$

This is indeed a very good introduction to Lambert function.

Sooner or later, you will learn that any equation which can write or rewrite $$A+Bx+C\log(D+Ex)=0$$ has solution(s) which espress(es) in terms of this function.

In the case of natural logarithms, using the steps shown in Ross Millikan's answer, you would end with $$x=12+\frac{1}{\log_e (2)}W\left(\frac{e^2 \log_e (2)}{4096}\right)$$

Assuming logarithms in base $2$ as the numbers suggest, then, as Ross Millikan answered, $$x=12+\frac{1}{\log_e (2)}W\left(\frac{\log_e (2)}{1024}\right)$$

Assuming logarithms in base $10$, $$x=12+\frac{1}{\log_e (2)}W\left(\frac{25 \log_e (2)}{1024}\right)$$

Now, since the argument is quite small, you can approximate the value of $W(t)$ using the expansion $$W(t)=t-t^2+\frac{3 }{2}t^3-\frac{8 }{3}t^4+O\left(t^5\right)$$ or, better, using Padé approximants such as $$W(t)=\frac{t }{1+t }$$ $$W(t)=\frac{t+\frac{4}{3} t^2}{1+\frac{7 }{3}t+\frac{5 }{6}t^2 }$$

Once you arrive at $100$ = ( $x−12$ )( $2^x$ ) ,

$x-12$ = $100$ * $(2^{-x})$ ,

x = 100*(2^(-x)) + 12 ,

On my calculator, to get an approximation, i just started with some random number ( $1$ in my case), and entered $100$ *( $2^{-ans}$ ) + $12$ , where ans is the previous value. By repeatedly evaluating this, it converges to to the value of $x$ = $12.0240111$ . In theory, this should work from any initial value.

1 $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g. , basic help on mathjax notation , mathjax tutorial and quick reference , main meta site math tutorial and equation editing how-to . $\endgroup$ – José Carlos Santos Oct 21, 2020 at 6:21

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Algebra
Section 6.4 : Solving Logarithm Equations · log5(2x+4)=2 log 5 ( 2 x + 4 ) = 2 · logx=1−log(x−3) log ⁡ x = 1 − log ⁡ ( x − 3 ) · log2(x2−6x)=3
Logarithmic Equations: Very Difficult Problems with Solutions
Logarithmic Equations: Very Difficult Problems with Solutions ; x y + y x = l o g 4 32 · +xy=log432 ; x y + y x = 1 2 l o g 2 32 · +xy=21log232 ; x 2 + y 2 x y
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Solving challenging logarithm equation
This needs a numeric solution and is in a good form as the right side will change slowly with y. Let y0=0 and iterate. After two iterations we