A Non-standard Logarithmic Equation with Different Bases

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[arundhatimukherjee]

a>-1 or a+8>7 or log (a+8) to the base 4 > log 7 to the base 4 (as log x to the base 4 is an increasing function) >1. So k>1. For the function f(x)=4^x-3^x-7, f'(x)= 4^x ln4 - 3^x ln3>0 (as 4^x>3^x>0 for all x>1 and ln4>ln3>0 and we can multiply the inequalities as all the terms are positive). So f(x) < 0 for all x in (1, 2) and >0 for all x>2. So a=8 is only solution. I did all these just to avoid calculating the critical point of the function.🙂

[RexxSchneider]

You don't make clear why when x<x0, the derivative of 4^x - 3^x must be decreasing. It's perfectly possible for a derivative to be zero at some point, but be increasing at either side (a point of inflection).
It should be sufficient to observe that the values of both 4^x and 3^x are between 0 and 1 when x is negative, so their difference can never be 7 for negative x.
Then it is clear that derivative of 4^x - 3^x is always positive for positive x (4^x > 3^x and ln4 > ln3), so the function is monotonically increasing, which shows that x=2 is the only value of x that can make the function equal to 7.
No calculators needed.

[alessandrorossi]

When you take the derivative of f(x) you can also just note that 4^x > 3^x and ln4 > ln3 to establish that f'(x) is always positive and therefore f(x) is monotonic so that f(x) = 7 will have at most one solution (in the reals).

[manojsurya]

It was a nice logarithmic equation I tried something with log_b (a) =ln(a)/ln(b) but couldn't proceed after some steps, Ur approach was good😌

[leecherlarry]

compi loved the problem, thanks!
*Solve[Log[3, a + 1] == Log[4, a + 8], a, Reals]*

[ashishpradhan]

Wonderful explanation with graph. ♥️♥️♥️

[haricharanbalasundaram]

I have a super difficult question, try this: if a, b, c are the sides of a triangle such that x^2 - 2 ( a + b + c) x + 3 k (ab + bc + ca) = 0 has real roots, find the range of k.

(Adapted from JEE advanced, not sure which year)
:D

[carloshuertas]

Another great explanation, SyberMath!

[aashsyed]

Awesome problem
I will keep supporting you !

[KJ-zspi]

Thanks man these type of problems are just cool

[sujanshankarbhowmick]

Pls, try this elegant problem. For the summation where n>1, how many +ve integral solutions for z are possible?

[wafiklotfallah]

You can also write:
f(x) = log(...) - log(...)
Check that f(8) = 0 and f'(x) > 0 whenever defined.

[yoav]

The derivative of log3(a+1) is 1/(ln3(a+1)) and the derivative of log4(a+8) is 1/(ln4(a+8)) so the derivative of log3(a+1) is always bigger than the derivative of log4(a+8) so once thees functions meet they cant meet again so x=8 is the only solution

[ramaprasadghosh]

log ( a+1) to the base 3
= log ( a+8) to the base 4
= log ( a+8) to the base 3
* log (3) to the base 4
or a +8
= (a+1)^ ( log 4 to the base 3)

[pokemil]

Consider equation 4^x = 3^x + 7
1 = (3/4)^x + 7/(4^x)
The function (3/4)^x is strictly decreasing (because 3/4<1), and so is the function 7/(4^x)=7*(1/4)^x (because 1/4<1). That means the function (3/4)^x + 7/(4^x) is also strictly decreasing, so there is at most one solution to this equation. We can check that x=2 is indeed a solution, so the only real solution to 4^x=3^x+7 is x=2.

One more thing, here is the proof to the statement that if f and g are strictly decreasing, then f+g is also strictly decreasing.
Consider any two real numbers x1<x2 => f(x1)>f(x2) and g(x1)>g(x2)
Then the function h(x)=f(x)+g(x) is strictly decreasing because

[satrajitghosh]

log(a+1) to the base 3
= log(a+8) to the base 4
= x, say
So a+1 = 3^x and a+8 = 4^x
So 4^x - 3^x = 7
This has a feasible solution at x= 2
Again 4^x - 3^x - 7 < 0 for x < 2
And 4^x - 3^x - 7 > 0 for x > 2
Hereby x=2 is the only fessible solution
Hereby a =3^x -1 = 8

[Raynover]

f is negative when x<0 (algebraically)
f is strictly increasing in [0, +oo)

[math]

very well done, explained perfectly bro

[meccamiles]

This channel is just good fun. Keep it up.

[CARLESIUS]

¿a =8? Logarithm to base 3 of 9 is 2 and logarithm to base 4 of 16 is 2.

The solution was as simple as finding that 3, 4, and 5 as lengths form a right triangle. A Pythagorean triple.