1

I am trying to resolve the equation $$\log_5 (2x+1) = \log_3 (3x-3)$$ and then of sketch the functions $y=\log_5 (2x+1)$ and $y=\log_3 (3x-3)$ get the solution $x=2$. There is an method that do not use the graphic method? Thanks for your suggestions. Any ideas will be appreciated.

Shaun
  • 44,997
Ricardo Largaespada
  • 680
  • 2
    $$\log_3(3x-3)\cdot\log_5(3)=\log_5(3x-3)$$ – Rushabh Mehta Dec 20 '18 at 19:10
  • 2
    You can hope that your teacher isn't out to torture you and $(2x+1) = 5^n$ and $(3x-3)= 3^n$ for some natural number $n.$ Start with, $0$, then try $1$ and find that a solution exists. After you find one solution, argue that both the LHS and RHS are upward sloping, but one is always more upward sloping than the other, so there can be no other intersection points. – Doug M Dec 20 '18 at 19:17
  • 1
    Hint: $$\log_a(x)=\frac{\log_b(x)}{\log_b(a)}.$$ (Well, that's where I'd start anyway . . . ) – Shaun Dec 20 '18 at 19:18

2 Answers2

2

Let $$f(x) = \log_5 (2x+1)-\log_3(3x-3)$$ which is defined for $x>1$.

Since for $x>1$ we have $$(x+{1\over 2})\ln 5 > (x-1)\ln5 >(x-1)\ln3>0$$

we have also that $$f'(x) = {1\over (x+{1\over 2})\ln 5 } - {1\over (x-1)\ln3} <0$$

so given function has at most one zero ...

nonuser
  • 90,026
  • That is at most we can do at such equation. We are left to guess a solution somehow. Any way you can also try with some aproximaton method. – nonuser Dec 20 '18 at 19:32
  • You right I already think about derivate the function $f(x)$ and verify that is monotonous. Then it has at most one zero. – Ricardo Largaespada Dec 20 '18 at 19:37
0

If you want a purely algebraic solution, although rather tedious, you can solve the equation as follows:

$$\log_5(2x+1) = \frac{\log_3(2x+1)}{\log_3 5} \tag{1}$$

$$\frac{\log_3(2x+1)}{\log_3 5} = \log_3(3x-3) \tag{2}$$

$$\left(3^{\log_3(2x+1)}\right)^{\frac{1}{\log_3 5}} = 3x-3 \tag{3}$$

$$(2x+1)^{\frac{1}{\log_3 5}} = 3x-3 \tag{4}$$

$$(2x+1)^{\log_5 3} = 3x-3 \tag{5}$$

$$\log_{2x+1} (3x-3) = \log_5 3 \tag{6}$$

$$\boxed{x = 2} \tag{7}$$

$(1)$: Change of bases: $\log_a b = \frac{\log_c b}{\log_c a}$

$(2)$: Rewriting the equation

$(3)$: Rewriting as an exponential: $c = \log_a b \iff a^c = b$

$(4)$: Simplifying

$(5)$: Inversion of base and argument: $\frac{1}{\log_a b} = \log_b a$

$(6)$: Rewriting as a logarithm: $a^b = c \iff \log_a b = c$

$(7)$: Solution to $2x+1 = 5$ and $3x-3 = 3$

But of course, you have to check for any extra solutions. Since $\log_a b^c = c\log_a b$ and $\log_{a^c} b = \frac{1}{c}\log_a b$, you have $$\log_{a^ c} b^c = \log_a b$$

$$\implies \log_5 3 = \log_{5^n} 3^n$$

$$\implies\log_{2x+1} (3x-3) = \log_{5^n} 3^n$$

$$\implies 2x+1 = 5^n; \quad 3x-3 = 3^n$$

$$3x-3 = 3^n \implies x-1 = 3^{n-1} \implies x = 3^{n-1}+1$$

Plugging this in the first equation, you have

$$2\left(3^{n-1}+1\right) = 5^n \implies 2\left(3^{n-1}\right)+3 = 5^n$$

It’s fairly obvious the only solution is $n = 1$ (since the RHS grows at a much faster rate), for which we already solved the equation. Hence, it’s the only solution.

KM101
  • 7,176