Log_0.1 OF x=2log_0.1 OF 6 - 0.5log_ 0.1 OF100+ 3log_0.1 OF ∛20?

If those are the log bases near the underscore, then I think you have:

log_0.1(x) = 2 log_0.1(6) - 0.5 log_0.1(100) + 3 log_0.1(∛20)

All logs have the same base, so let's start out with making all three terms on the right side all under a single log.

First, the numbers outside the log can be brought inside as an exponent:

log_0.1(x) = log_0.1(6²) - log_0.1(√100) + log_0.1(∛20³)

simplify:

log_0.1(x) = log_0.1(36) - log_0.1(10) + log_0.1(20)

Now the difference of two logs is the same as the log of the quotient and the sum of two logs is the same as the log of the product. So the addition and subtraction of logs can be changed into a single log:

log_0.1(x) = log_0.1(36 * 20 / 10)

we can simplify that:

log_0.1(x) = log_0.1(36 * 2)

log_0.1(x) = log_0.1(72)

And finally, we have two values equal to each other with the same base's logs on each side. So what's inside the logs must also be equal, so:

x = 72

As a test, if we substitute that x into the original equation, use the base change rule to get decimal approximations, we should get close to the same value on both sides (won't be exact due to rounding, but will be close), so we have:

log_0.1(x) = 2 log_0.1(6) - 0.5 log_0.1(100) + 3 log_0.1(∛20)

log_0.1(72) = 2 log_0.1(6) - 0.5 log_0.1(100) + 3 log_0.1(∛20)

log(72) / log(0.1) = 2 log(6) / log(0.1) - 0.5 log(100) / log(0.1) + 3 log(∛20) / log(0.1)

log(0.1) = -1, so make that substitution:

-log(72) = -2 log(6) + 0.5 log(100) - 3 log(∛20)

log(100) = 2, so make that substitution:

-log(72) = -2 log(6) + 0.5(2) - 3 log(∛20)

-log(72) = -2 log(6) + 1 - 3 log(∛20)

Now let's use a calculator for the rest:

-1.8573325 = -2(0.7781513) + 1 - 3 log(2.7144176)

-1.8573325 = -2(0.7781513) + 1 - 3(0.4336767)

-1.8573325 = -1.5563026 + 1 - 1.3010301

-1.8573325 = -1.8573327

Again, not exact, but close enough to believe our answer is correct.

Again, the answer is:

x = 72

x = 36 / 10 + 20 = 23.6