I need help with these math problems (Not logarithms) Thanks 5^x = 1/25?

For 1 and 2 make the bases equal, then that sets the exponents equal.

1] 5^x = 1/25

5^x = 1 / 5^2

5^x = 5^-2

So x = -2

2] 4^(11- 3x) = 16^x

4^(11- 3x) = 4^2^x

4^(11- 3x) = 4^2x

So 11- 3x = 2x

11 = 5x and 11/5 = x

3] is missing something because 2^2 = 4 not 20

4} Think like this, The log is the exponent.

So log(5) = x says the log is x. (the exponent is x)

The base is 5 because it says so.

And so far you have a base and an exponent. Namely 5^x.

The 12 is left over so you write 5^x = 12.

This not the correct mathematics terminology but rather a quick fix that is still leads to the correct answer.

1.) 5^x = 1/25 = 1/5^2 = 5^-2

so x = -2

2.) 4^(11-3x) = 16^x (4^2)^x = 4^2x

so 11- 3x = 2x

5x = 11

x = 2 1/11

(Logarthims problems)- Im having trouble with?

3.) 2^x = 20 ?

take logs of both sides

x log 2 = log 20

x = log 20 / log 2 = you calculate

4.) log(5) 12 = x

x = log 12 / log 5 = you calculate

/// Rule; loga b = log b / log a

Ok I tried this for fun because I'm lame but I got

1.)5x= 1/25 > 1/25 (5)= X

X=0.2

I didnt do 2.)

3.) 2^2 =20 > 20/2^2 =10

Assuming 2^2= two exponent two

4.) log(5) 12 = 0.416 <if

You divide long(5) by 12

Or

log(5) 12 = 60 <if you times log(5) by 12

Did this for fun don't hate

Question 1)

Before doing anything, the bases need to be the same. 5 and 1/25 have a common base of 5. So change 1/25 to 5^-2:

5^x = 5^-2

Now that the bases are the same, drop the powers and put them into an equation:

x = -2

———–––—–

Question 2)

Once again the bases need to be the same, so change 16 to 4^2:

4^(11-3x) = (4^2)^x

To have it all as one power, multiply the power inside the brackets by the power outside the brackets:

4^(11-3x) = 4^2x

Now that the bases are the same, drop the powers and put them into an equation:

11 - 3x = 2x

11 = 5x

x = 11/5

———————

Question 3) (assuming you meant 2^2x = 20)

2^2x = 20

As 2 and 20 do not have a common base, take the log of both sides:

log(2^2x) = log20

Now divide both sides by log2:

2x = log20/log2

Divide both sides by 2 by multiplying straight into the denominator:

x = log20 / 2log2

That is your solution in its exact form, for a decimal approximation put that into a calculator:

x ≈ 2.160964047

———————

Question 4)

As there is a base of 5 and an exponent of x, change the equation so that it fits:

5^x = 12

As 5 and 12 do not have a common base, take the log of both sides:

log(5^x) = log12

Divide both sides by log5:

x = log12/log5

That is your answer in its exact form, for a decimal approximation put that into a calculator:

x ≈ 1.543959311

For 1 and 2 what are you exactly looking for? What did the question said solve for x? The question is not clear therefore can't help. Double check the question and then post it again.

3.) 2^2 = 20

2 = log2 20

4.) log(5) 12 = x

5^x = 12

1. 5^x = 1/25

5^x = 5^-2

x = -2

2. 4^(11-3x) = 16^x

4^(11-3x) = 4^2x

11-3x = 2x

5x = 11

x = 11/5

3. 2^2 = 20, Not true?

4. log(5) 12 is approximately 1.54

1) 5^x = 1/25 5^x = 1/5^2 5^x = 5^-2 x = -2 2) 4^(11-3x) = 16^x 4^(11-3x) = (4^2)^x 4^(11-3x) = (4)^2x 11-3x = 2x 11 = 5x x = 11/5 3.) 2^2 = 20 this is not the problem !!!...possible it was 2^x = 20 ? 2^x = 20 x ln(2) = ln(20) x = ln(20)/ln(2) x =4.32 4.) log(5) 12 = x use the change of base function ln(12)/ln(5) x = 1.543