Not sure how to find the exponent X in equation.

[SashaK]

4^3x+1 = 2 * 9^3x+0.5
here's what I have so far:
(2^6x+2) = 2 * (3^6x+1)
(2^6x) * 4 = 2 * (3^6x) * 3
(2^6x) * 4 = (3^6x) * 6

The answer in my book is (-1/6).
Thank you so much in advance!

I just spied another question that I don't understand on my page:
3^2x-1 = (1/5)^2x-1

 

[soroban]

Hello, SashaK

Here is a theorem you may not be aware of . . .

\(\displaystyle \text{If }a^n \:=\:b^n\,\text{ and }\,a \ne b,\,\text{ then: }\,n = 0\) . **

\(\displaystyle 4^{3x+1} \:=\: 2\cdot9^{3x+0.5}\)

\(\displaystyle \text{We have: }\;\left(2^2)^{3x+1} \:=\:2\cdotleft(3^2\right)^{3x + 0.5} \quad\Rightarrow\quad 2^{6x+2} \:=\:2\cdot3^{6x+1}\)

\(\displaystyle \text{Divide by 2: }\;2^{6x+1} \;=\;3^{6x+1}\)

\(\displaystyle \text{The theorem says: }\:6x+1 \:=\:0 \quad\Rightarrow\quad x \:=\:-\frac{1}{6}\)

\(\displaystyle 3^{2x-1} \:=\: \left(\tfrac{1}{5}\right)^{2x-1}\)

\(\displaystyle \text{The theorem says: }\:2x-1 \:=\:0 \quad\Rightarrow\quad x \:=\:\frac{1}{2}\)


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**

\(\displaystyle \text{We have: }\:a^n \:=\:b^n\;\text{ and }a \ne b.\)

\(\displaystyle \text{Take logs: }\;\ln(a^n) \:=\:\ln(b^n) \quad\Rightarrow\quad n\cdot \ln a \:=\:n\cdot \ln b\)

. . \(\displaystyle n\cdot\ln a - n\cdot\ln b \:=\:0 \quad\Rightarrow\quad n\cdot(\ln a - \ln b) \:=\:0\)

\(\displaystyle \text{Hence, we have: }\:n \:=\:0\;\;\text{ or }\;\ln a \:=\:\ln b\)


\(\displaystyle \text{Since }a \ne b,\,\text{ then: }\,\ln a \ne \ln b\)

. . \(\displaystyle \text{Therefore: }\:n \:=\:0\)

 

[galactus]

Cool, Soroban.

 

[mmm4444bot]

For those who don't know the above theorem, apply basic properties of logarithms to the given equation and simplify.

Doing that yields:

(6x + 1) * log(2) = (6x + 1) * log(3)

We see by inspection that the resulting equation is true only when 6x + 1 = 0.