inequality with log and expo func.

[runic eagle]

Hello, how to approach an inequality with both an exponential function and logarithm. See the picture. Thanks in advance.

[woven creek]

This is a product

You can study the factors separately

When is 2^x >0?

When is log(3x-1)>0?

Also remember that they could be both negative too make this a true statement

[runic eagle]

My bad for the blue part, x > 0

[woven creek]

Do you know what 2^(-1) is?

[runic eagle]

2

1/2

[woven creek]

Which is still positive!

Also there's a mistake in the last line of the yellow

[runic eagle]

Yeah 2/3 sorry im a mess

[runic eagle]

But why did you choose -1?

[woven creek]

It's okay to be a mess sometimes

Just to get you thinking

[runic eagle]

Okay thanks a lot

But answering your question

When is 2^x > 0

I’ve learnt the rule
g^a = b <-> a = log_g(b)

So,

2^x > 0 <-> x = log_2(0)

Which should be undefined

However

If I take look at the inequality, any positive number for x, in 2^x will be automatically > 0

Even if x = 0

Even x = -10

So how should I approach them?

[woven creek]

Is there any x that makes the inequality false?

[velvet tundra]

really simply:
whenever you have 2 (or more) numbers that when multiplied are =, > or < 0:
simply solve for when each of them is 0
those are all of your possible values (union not intersect, because as long as one of them is 0, the entire product will be 0)
also, an exponential with positive base will always be > 0, so 2^x will never be 0, so you can ignore that part and just work out the log part)

if it helps, draw a graph

[runic eagle]

Thanks I got💯🔥

[runic eagle]

Dude, legend. This helped so much. I solved it within 20 seconds. ‼️🔥

Thanks y’all 💯

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