There are several possible interpretations for the given equation. One might be
[tex]3^{6x+4} = 4^{x+8}[/tex]
Rewrite the left side as
[tex]3^{6x+4} = \dfrac{3^{6x+48}}{3^{44}} = \dfrac{(3^6)^{x+8}}{3^{44}}[/tex]
Then in the equation, dividing both sides by [tex]4^{x+8}[/tex] gives
[tex]\dfrac{(3^6)^{x+8}}{3^{44} \cdot 4^{x+8}} = \dfrac1{3^{44}} \left(\dfrac{3^6}4\right)^{x+8} = 1[/tex]
Solve for [tex]x[/tex].
[tex]\left(\dfrac{3^6}4\right)^{x+8} = 3^{44}[/tex]
[tex]\log_{3^6/4}\left(\left(\dfrac{3^6}4\right)^{x+8}\right) = \log_{3^6/4} \left(3^{44}\right)[/tex]
[tex]x+8 = \log_{3^6/4} \left(3^{44}\right)[/tex]
[tex]x = \log_{3^6/4} \left(3^{44}\right) - 8 = \boxed{44 \log_{729/4} (3) - 8}[/tex]
Answer:
[tex]x=\dfrac{8\log_32-2}{3-\log_32}[/tex]
Step-by-step explanation:
Given equation:
[tex]3^{6x+4}=4^{x+8}[/tex]
Rewrite 4 as 2²:
[tex]\implies 3^{6x+4}=(2^2)^{x+8}[/tex]
[tex]\textsf{Apply the exponent rule} \quad (a^b)^c=a^{bc}:[/tex]
[tex]\implies 3^{6x+4}=2^{2x+16}[/tex]
Take log with base 3 of both sides:
[tex]\implies \log_33^{6x+4}=\log_32^{2x+16}[/tex]
[tex]\textsf{Apply the Power law}: \quad \log_ax^n=n\log_ax[/tex]
[tex]\implies (6x+4)\log_33=(2x+16)\log_32[/tex]
[tex]\textsf{Apply the log law}: \quad \log_aa=1[/tex]
[tex]\implies 6x+4=(2x+16)\log_32[/tex]
Expand the brackets:
[tex]\implies 6x+4=2x \log_32+16 \log_32[/tex]
Collect like terms:
[tex]\implies 6x-2x \log_32=16 \log_32-4[/tex]
Factor:
[tex]\implies 2x(3-\log_32)=2(8\log_32-2)[/tex]
[tex]\implies x(3-\log_32)=8\log_32-2[/tex]
Isolate x:
[tex]\implies x=\dfrac{8\log_32-2}{3-\log_32}[/tex]
Therefore, x as a decimal is:
[tex]\implies x=1.286343465...[/tex]
Learn more about log laws here:
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