Can anyone help to solve this? (Indices and logarithms)

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[Sim Thiam Huat]

Solve the equation
3^2x+1-2(3^x+0.5)-3=0

the answer given is 0.5 but i can't get it. Just hope anyone here can help?

[Sim Kwang Yaw]

The equation can be simplified to 3(3^2x) - 2 × sq rt(3)3^x – 3 = 0
Solving the equation 3y² - 2sqrt(3)y - 3 = 0 should give answer sqrt 3 or -1/3 sqrt(3)

So I suppose o.5 is not the answer. Answer should be 3^0.5

[Chang T H]

3^x = 3^0.5 or -3^(-1.5).
But 3^x is > or = 0.
So the negative answer is not acceptable.
Thus, 3^x = 3^0.5.
Or, x = 0.5.

[Sim Kwang Yaw]

Thanks Tze Hin, I forgot the last step. y = 3^x = 3^0.5

Apology for overlooking.

[Goh Ching Huwat]

Solve the equation
3^2x+1-2(3^x+0.5)-3=0

the answer given is 0.5 but i can't get it. Just hope anyone here can help?

to solve this problem, just let y=3^x+0.5)
the equation become y²)-2y-3=0
use simutanuous equation.
that's all

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