Solve the equation, 9^4x-3=27^5x
\[9^4x-3=27^5x\] is this ur ques?
I think it is: \[\Large {9^{\left( {4x - 3} \right)}} = {27^{5x}}\] @imqwerty
ah yes :)
In the first part, the x-3 is in with the power, and in the second part the x is also in with the power. Yes, Michele_Laino that's right
\[9^{4x-3}=27^{5x}\]\[3^{2(4x-3)}=3^{3(5x)}\]when bases are same powers are equal so 2(4x-3)=5x can u get x from here?
So now do you just solve for x?
yes!
Okay.
Does x=2?
hint: we can simplify the equation above like below: \[\Large \begin{gathered} 2\left( {4x - 3} \right) = 3\left( {5x} \right) \hfill \\ 8x - 6 = 15x \hfill \\ \end{gathered} \]
:)
So the answer is -6/7?
correct!
Okay, so question with the hint that you gave me where did you get the 3(5x) from?
Okay, so question with the hint that you gave me where did you get the 3(5x) from?
since I can write this: \[\huge {27^{5x}} = {\left( {{3^3}} \right)^{5x}} = {3^{\left( {3 \cdot 5x} \right)}}\]
Ohh, okay. Got it now. Thanks so much!!
:)
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