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

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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....Show more

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...Show more

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...Show more

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...Show more

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...Show more