How do I solve this logarithmic problem?

Question

Write the expression using only natural logarithms:

log_(1/6)_6x^2

The _'s indicate that 1/6 is the base.

Answer 1

So you want to solve

log [base 1/6] (6x^2).

And you want to solve it using only natural logarithms, or log[base e], or ln.

The first thing to note is that, using the log property
log[base b](ac) = log[base b](a) + log[base b](c)

we can change the form of that log to

log[base 1/6](6) + log[base 1/6](x^2)

Also, using the log property that
log [base b](a^c) = c * log[base b](a),

log[base 1/6](6) + 2 log[base 1/6](x)

One thing to note is that if we let y = log[base 1/6](6), we can actually reduce the log itself.

y = log[base 1/6](6). Changing this into exponential form, we get

(1/6)^y = 6, or
( [6]^(-1) )^y = 6, or
6^(-y) = 6^1

So we can equate the exponents to get
-y = 1, or
y = -1.

So log[base 1/6](6) = -1. Substituting this in, we get

-1 + 2 log[base 1/6](x)


Now, we use the change of base formula to convert our current logs into natural logs, ln. Note that the change of base formula goes as follows:

log[base c](a) = log[base b](a) / log[base b](c), where we can choose b to be *any* base.

In this case we want b to be e, so we'll end up with ln.


-1 + 2 log[base 1/6](x)
-1 + 2 {ln (x)} / {ln(1/6)}

Note that ln(1/6) = ln(6^(-1)) = -ln(6), so we can have

-1 + 2 [ln(x)] / [-ln(6)], or

-1 - 2 ln(x)/ln(6)

Answer 2

Now that's a weird base. 1/6. As an enumeration base, it's a little unusual. For example, thirty-six is 0.01 in that base. Now if it were 1/7, since 7 is prime, you would get a really interesting notation, completely redefining distance, by using powers of 7; something called 7-adic numbers.

But that's another topic. To solve this, use the identity:

log(a)b = log(x)b/log(x)a,

where a, b, and x are any numbers that can be used as bases for logarithms (i.e., don't use 0, 1, or negative numbers).

In your case,

log(1/6)6x^2 = log(1/6)6 + 2 log(1/6)x
= log(6)(6)/log(6)(1/6) + 2 log(e)(x)/log(e)(1/6)
= 1/(-1) + 2 ln(x)(-ln 6) = -(2/ln6) ln(x) - 1

Answer 3

for the logarithm of a nunber in other bases, we have the formula

log_a_(b) = log(b)/log(a) = ln(b)/ln(a)

where a is the base and b is the argument of the logarithm

so if we used this in your expression

log_(1/6)_6*x^2 with a = 1/6 and b = 6*x^2

log_(1/6)_6*x^2 = ln(6*x^2)/ln(1/6) = 2*ln(6*x)/ln(1/6)

Answer 4

Could you rephrase your question or something. Natural logarithms are base e. You say the base of this problem is (1/6).

Answer 5

By the change of base formula log_b_(x) = ln(x)/ln(b)

So it's ln(6x^2) / ln(1/6)

Answer 6

log_(1/6)_6x^2=ln 6x^2 *ln 1/6
ln 6x^2 *ln 1/6

Answer 7

ln(6x^2)/ln(1/6)
=(ln6 + 2lnx)/(-ln6)
=-1 - 2 lnx/ln6