# i want to know what x is in the logaritme below: 3log x + 2 log 3x= 3 the answer is 3 but how do you get this answer thank you

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3log x + 2 log 3x = 3

First we know that:

a log b = log b^a

==> 3 log x + 2 log 3x = 3

==> log x^3 + log (3x)^2 = 3

==> log x^3 + log 9x^2 = 3

Also, we know that:

log a + log b = log a*b

==> log (x^3)*(9x^2) = 3

==> log 9x^5 = 3

==> 9x^5 = 10^3

==> x^5 = 10^3/9

==> x = (1000/9)^1/5

First, we'll have to precise that we work with decimal logarithms.

Now, we'll use the power property of logarithms:

3log x = log x^3

2 log 3x = log (3x)^2

Now, we'll re-write the equation:

log x^3 + log (3x)^2 = 3

Because the bases of logarithms are matching, we'll use the product property of logarithms: the sum of logarithms is the logarithm of the product.

log x^3 + log (3x)^2 = log [x^3*(3x)^2]

log [x^3*(3x)^2] = 3

We could write the term from the right side as:

3 = 3*1 = 3* log10

The equation will become:

log [x^3*(3x)^2] = 3* log10

log [x^3*(3x)^2] = log10^3

Because the bases are matching, we'll use the one to one property:

x^3*(3x)^2 = 1000

9*x^(3+2) = 1000

9x^5 = 1000

We'll divide by 9:

x^5 = 1000/9

x^5 = 111.111 approx.

x = (111.111)^1/5 approx.

To solve for x in the equation 3logx +2log3x = 3

solution:

By property og logarithms,

loga*b = log a+logb

log a/b = log a- logb

m log a = log a^m.

Using the above , the second term 2log3x = 2 (log3+logx) = 2log2+ 2logx. Now we rewrite the given equation:

3logx +2log3+2logx = 3.

3logx+2logx = 3-2log 3.

5 logx = log10^3 - log3^2, as 3 = log10^3

5logx = log (10^3/log3^2), as loga-b = log (a/b).

logx = (1/5)log (1000/9).

logx = log (1000/3)^(1/5). Take antilogarithms:

x = (1000/9)^(1/5) = (111.111..)^(1/5) = is a number between 2 and 3 as 2^2 = 32 and 3^5 = 243.

x = (111.11...)^(0.2) = 2.56537878 by excel