how to solve logarithmic equations with different bases

Solving logarithmic equations with different bases However, I do not know the approach to getting the solution to one of them.

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#math #algebra #mathbasics #logarithm

[stevemonkey]

It says something good about you that you posted a problem that you cannot solve. But I cannot solve it either. 😔

[michaelempeigne]

log_2 ( x ) = log_3 ( x + 1 ) = y

so 2^y = x
and 3^y = x + 1
3^y = (2^y ) + 1
3^y - 2^y = 1
this is easy to tell that y = 1

that means log_2 ( x ) = 1
so x = 2

[EE-hoiz]

My solution is as follow:

log_2(x) = log_3(x+1)
= ln(x)/ln(2) = ln(x+1)/ln(3) (as shown in the video)

Here's where I deviate from Steve's path: I multiply both side by ln(2)/ln(x+1), which yields:
ln(x)/ln(x+1) = ln(2)/ln(3)

By comparison, x=2 and x+1=3, which of course still yields x=2

Therefore x=2. Voila!!!


Edit: An honorable mention to Chirag Ahuja in the comment section, who came up with e^ln(x) = e^ln(2)log_3(x + 1) through some manipulation, and had the idea of "compare ln(x) with ln(2) and log_3(x + 1) with 1". It also used the method of solving by comparison.
It's so interesting that I can't help but mentioning it!

[chiragahuja]

So these can be the possible explanations:

1. So I ended up with a equation after some manipulations
e^ln(x) = e^ln(2)log_3(x + 1)
my thought is how about after equating the exponents since bases are same we compare ln(x) with ln(2) and log_3(x + 1) with 1 this works out both yields the ans as 2 and makes sense as well. It might be wrong but makes sense for me.

2. another expression is ln(3)^ln(x) = ln(x+1)^ln(2) now we can just compare ( ln(3) = ln(x+1) ^ ln(x) = ln(2) ) and we end up with 2

3. One more I came up with is:
Log_(x+1)(x) = log_3(2)
Am thinking of comparing bases and the number which we are taking log of but this just seems wrong but on a second thought I thought I should still put

[moskthinks]

I don't know how to solve the case for different bases, but I can show this equation only has x=2 as its answer. Note that it satisfies the equation.

Then define f(x)=log2(x)-log3(x+1)
We have f'(x) = 1/(xln2)-1/((x+1)ln3)
=

So for the domain of f(x) [x>0], we have f'(x)>0, which means f is strictly increasing. Because of that, f(x) must have only 1 root, which happens to be x=2. QED.

Although now what😅

[WolfgangKais]

Transforming the second equation to
ln(x+1)/ln(x) = ln(3)/ln(2) helps to „see“ the solution, and showing that the LHS is decreasing (for x>1) ensures that there is only one solution. I was able the solve the related equathions when replacing the log base 3 by base 4 or base 8, and I am pretty sure that there is no solution when replacing base 2 by base 5.

[namelessnormie]

so here is a simple mathematical approach to solve:
log_(2)x=log_(3)(x+1)

1) use change of base rule(as mr BPRP)
(lnx )/(ln2) = (ln(x+1))/(ln3)

then rewrite the equation:- (lnx )/(ln(x+1))=(ln2)/(ln3)
now we know two things
1) lnx=ln2 and
2) ln(x+1)=ln(3)
we know log_(a)b = log_(a)c if and only b=c
so based on this x=2 and x+1=3

but now the problem is this approach will not work if the question was
log_(2)x= log_(5)(x+1)

[padraiggluck]

For the second problem, multiply the first line in blue on both sides by ln(3)/ln(x) to get ln(3)/ln(2) = ln(x + 1)/ln(x) and the answer x = 2 is obvious.

[nicolascalandruccio]

In fact, most cases are not explicitely solvable.
The general case is:
log_a(x)= log_b(x+1) where, x>0, a>1, b>1 and b>a
Let t=ln(b)/ln(a)
Thus, t>1
ln(x)/ln(2)=ln(x+1)/ln(3)
(ln(3)/ln(2)).ln(x)=ln(x+1)
t.ln(x)=ln(x+1)
ln(x^t)=ln(x+1)
x^t=x+1
x^t-x-1=0
Therefore, there’s no explicit solution even using Lambert W function except in some rare cases where a is a natural number and b=a+1 solvable by comparison and the solution is x=a as @Chirag Ahujalike said for the case where a=2 and b=3 or some polynomial cases where t=2, t=3 or t=4.
For instance, the solution for t=2 is the golden ratio and for t=3,

[serulu]

Thankfully i solved this despite not learning log in school.

Log really seems easy, it's rules make a lot sense

[neomooooo]

So we’re looking at log_a(x)=log_b(x+c), as a general rule. Or, if we change notation,

m*ln(x)=ln(x+c), with m=ln(b)/ln(a).

This leads to the nonlinear equation

x^m=x+c,

which will only have a positive real solution if either m>1 and c>0, (a line with a positive intercept intersecting a concave polynomial through the origin), or when m<1 and c<0 (a line with a negative intercept crossing a convex function).

Only when m is rational is there any hope of getting a closed form answer, which would mean the bases are related via a=b^k, for some integer k. This is not the case when a=2, b=3. It looks like you’d have to resort to numerical approximation, e. g. Newton’s method.

With an initial guess of say x0=1, the next guess is

x1= x0 - (x0^m -x0 -c)/(m*x0^(m-1) -1),
=1+c/(m-1),

which will be a better estimate than x0. Then repeat n times,

…

x(n+1) = xn - (xn^m-xn-c)/(m*xn^(m-1)-1).

Newtons method is remarkably efficient: if the n-th estimate is correct to say 5 digits, the next estimate will be accurate to 10 digits, and the one after that to 20 digits.

[tbg-brawlstars]

Let base be a after we use base change property
Then 1st answer is
x = a^[(log3log2)/(log3log2)] all logs to base a
2nd one was easier than one
Here's how to do it
Logx/log2 = log(x+1)/log3
Using base change property reverse....
Log3/log2 = log 3 base 2 and log (x+1) /logx = log(x+1) base x
Log3 base 2 = Log (x+1) base x
X+1 = 3
X= 2
By comparing eqn. x=2

[tmsniper]

here's what i thought:
ln(x)/ln(2)=ln(x+1)/ln(3)
which gives ln(x)/ln(x+1)=ln(2)/ln(3)
if we call the function on the left side f(x)=ln(x)/ln(x+1)
then we are looking for solutions of f(x)=f(2), it's sufficient to show that f is one to one in order to prove that x=2 is the unique solution,
for all x>0
so we can safely say that f is increasing, therefore it is one to one so x=2 is the unique solution.

[jessetrevena]

I love you videos!! I watch every time you upload :)

[tsa_gamer]

Log(base2)X=Log(base(2+1))x+1
By comparison x=2

[julienmcat]

Here’s the solution to the second question:



log2(x) = log3(x+1)


log(x) / log(2) = log(x+1) / log(3)


Use cross substitution:



log(x) / log(x+1) = log(2) / log(3)


According to the equation, the numerators and denominators of both sides equal each other. Use the numerators to solve for x:


log(x) = log(2)

Therefore x = 2

You can also use the denominators to solve for x:

log(x+1) = log(3)

Therefore x = 2

[OptimusPhillip]

First equation: x = e^(1/ln(3/2)) [Well, I got my properties of logs mixed up. Somehow, I got the idea that 1/ln(a) = -ln(a)]
Second equation: I would know how to solve it if there was an identity for products/quotients of natural logs. As it sits, I'm stuck at ln(x)/ln(x+1) = 1/(ln(2)*ln(3)) [How the hell did I forget that ln(2) was in the denominator of the initial expression? I was on the verge of greatness, I was this close!]

[lexiette]

I think I know how to prove the latter equation, rewriting the original to: lg(x) / lg(x+1) = lg2 / lg3, you can then let 2 be y instead and now you have the same on both sides except for that x is y, and since its all in logs that means x=y!

[juliang]

Ln(x)/ln2=ln(x+1)/ln3
Ln(x)*ln(3)/ln(2) =ln(x+1)
Ln3/ln2=a. Note a>1
x^a=x+1
F(x)=x^a-x-1
F'=x^(a-1)-1=0 at x=1
F''=x^(a-2)>0 at x=1 therefore x=1 is the global maximum of F(x) (F'=0 has only one solution).
To find root analytically i have no idea, thought there might be smth here.

[BoagoGopadile]

Solve log2(27)=x²
—
log81(8)