Solve the following logarithmic problem for x in base 10 log base 10 (x+21) +log base 10 (x)= 2?

You don't have to write out "base 10". When no base is given for a log, it's assumed that the base is 10.

Remember that log(a) + log(b) = log(ab) for logs of any same base. So this means log[ (x+21)x ] = 2, which means (x+21)x = 10^2. This means x^2 + 21x - 100 = 0. Solve this for x.

Once you get your solutions, check them against the original equation. Remember you can't take the log of a number unless it's positive!

log (x + 21) + log (x) = 2

Recall your rules for logarithms, log(a) + log(b) = log(ab)

log(a) - log(b) = log(a/b)

Using this, we can combine the left side to

log [ (x + 21)(x) ] = 2

log (x² + 21x) = 2

Now we need to remove the log on the left. Recall that inverse functions "undo" each other. So if we can find the inverse of log(x), we can apply this function to both sides of the equation to isolate x² + 21x on the left. You may know that the inverse of log(x) is 10˟ . So if we apply the function 10˟ to both sides of the equation, we can get rid of the log on the left. On the right, we'll get 10², which we know how to deal with.

10^log(x² + 21x) = 10²

Recall that 10^log(stuff) = stuff. So,

x² + 21x = 10²

10² = 100, so

x² + 21x = 100

Now we can subtract 100 from both sides to obtain the quadratic:

x² + 21x - 100 = 0

This can be factored easily (review your factoring).

(x + 25)(x - 4) = 0

So x = -25, x = 4.

Now we need to make sure that both of these values can go into the functions we originally had (ie the x values are in the domain of each log (x + 21) and log (x).

Recall that the domain of log(x) = [0,∞), or all positive numbers.

If x = -25, x + 21 = -25 + 21 = -4

Since x + 21 is negative, -25 cannot be put into the log function, so x = -25 is not a solution.

Likewise, checking x = 4.

4 + 21 = 25, which is positive, so it's good. It also can go into log (x) since 4 is positive.

So we can come to a final answer of x = 4

Hi:

log( x+21) + log(x) = 2 - original equation

10^(log(x+21) + 10^(log(x) = 10^2 - using a exponent ( or anti log of equal base to both side of the equation to remove remove the logarithm.)

(x+21)(x) = 100 - solve the power or doing the anti log

x^2 + 21x = 100 - Multiplication

take 1/2 of the 21 which equal 10.5 and squaring than adding it to both side of the equation to complete the square

x^2 + 21x + 110.25 = 100+ 110.25 - De foiling or completing the square

(x+10.5)^2 = 210,25 - De foiling or square rooting the x^2 + 21x + 110.25

x = -10.5 +/- sq rt (210.25) - take the square root of both sides of the equation to remove a square

Answer # 1 :

x = -10.5 + 14.5 - Solving the square root

x = 4 - Addition

Answer # 2:

x = -10.5 - 14.5 - Solving the square root

x = - 25 - Addition

Since we are dealing with logarithms the Solution or answer is 4 because we can't take a negative logarithm

proof :

log( x+21) + log(x) = 2 - original equation

log( 4 +21) + log(4) = 2 - plug x with 4

log(25) + log( 4) = 2 - Addition

1.397940008672+ 0.602059991328 = 2 - solve the logs of 4 and 25

2 = 2 - Addition

it checks and equals

I hope this helps

log(x+21) + log(x) = 2

x(x+21) = 100

x² + 21x - 100 = 0

x = (-21±29)/2 = -25, 4

The negative result can be discarded since the logarithm function isn't defined for values less than or equal to zero.

x = 4

...

log base 10(x(x + 21)) = 2

100 = x² + 21x

x² + 21x - 100 =0

(-21 +/- √21² - 4(-100))/2 = x

(-21 +/- √841)/2 = x

(-21 +/- 29)/2 = x

x = 4

x = -25 (not possible)

log(x) ≡ log10(x)

... log(x+21)+ log(x) = 2

or log(x+21)+ log(x) = log(100)

or log( (x+21)*(x) ) = log(100)

or (x+21)*(x) = 100

or x² + 21x - 100 = 0

or (x + 25)(x - 4) = 0

or x = -25 ← ignore, no such thing as log(-25)

or x = 4 ← final answer