Question

Help me...

Answer 1

\[\huge \bf 2^{x-2} \times 3^x=4^x \times 9^{x+1}\]
Find the value of x

Answer 2

\[2^{x - 2}(3^x) = 4^x(9^{x + 1})\]

Expand using the rule
\(a^{b + c} = a^ba^c\)

\[2^x2^{-2}3^x = 4^x9^x9^{1}\]

Re-write as \[\frac{2^x3^x}{2^2} = 4^x9^x \times 9\]

Place like terms on the same side:
\[\frac{2^x3^x}{4^x9^x} = 2^2 \times 9\]

Answer 3

Utilize another rule:
\(a^cb^c = (ab)^c\) to re-write the left hand side as:

\(\dfrac{(2 \dot\ 3)^x}{(4 \dot\ 9)^x} = 4 \times 9\)

Answer 4

Furthermore utilize the rule

\(\frac{a^x}{b^x} = \left(\frac{a}{b}\right)^x\)

Answer 5

\(\left(\dfrac{6}{36}\right)^x = 36\)

Answer 6

Simplify the fraction then take logs of both sides.

Answer 7

then,
\[\huge \bf \frac{6^x}{36^x}=(\frac{6}{36})^x=36\]
\[\huge \bf 6^{-x}=6^2\]
so x=-2

Answer 8

correct/... @Hero

Answer 9

I wouldn't write it in that form.

Answer 10

The correct thing to do is isolate x properly

Answer 11

@Hero what is the meaning of this :-
\[\huge \bf n \in I^+\]

Answer 12

\[\left(\frac{1}{6}\right)^x = 36\]
\[x \log(1/6) = \log(6^2)\]
\[x = \frac{2 \log(6)}{\log(1/6)}\]
\[x = \frac{2 \log(6)}{\log(1) - \log(6)}\]
\[x = -\frac{2 \log(6)}{\log(6)}\]

Answer 13

Means n is an element of positive integers

Answer 14

oh!!! thank you so much for your help