Rewrite xlog2 + 3log2 - 2log3 in exponential form?

Answer 1

log2^x+log2^3-log3^2=log(2^(x+3))-log9=log[(2^(x+3))/9]

Answer 2

I think you need an equation if you want to write it in exponential form, but I'll do part of the simplification:
xlog2 + 3log2 - 2log3
=log2^x + log2^3 - log3^2
=log[2^(x+3)/9]

Answer 3

The three log rules used to combine/separate logs:

(1) log[base b](a) + log[base b](c) = log[base b](ac)
The sum of two logs is the single log of a product.

(2) log[base b](a) - log[base b](c) = log[base b] (a/c)
The difference of two logs is a single log of a quotient.

(3) log[base b](a^c) = c * log[base b](a)
Whenever you have an exponent within a log, it can be brought out outside of the log.

Remember that these rules also work in reverse.

Now we apply them.

First, move everything in front of the log as an exponent inside a log. So

x log(2) + 3log(2) - 2log(3) becomes

log(2^x) + log(2^3) - log(3^2), which we can simplify to
log(2^x) + log(8) - log(9)

Now, we use the log properties. First, the addition, to combine the first two logs into a single log.

log ( (2^x)(8) ) - log(9)

Note that 8 = 2^3

log ( (2^x)(2^3) ) - log(9)

And whenever we have the same base but different exponents, we can add them together.

log (2^(x + 3)) - log(9)

And now we use the difference rule of logs.

log ( [2^(x+3)]/9 )

Unfortunately, I cannot write that in exponential form because I need it equal to something. If, however, you meant

y = xlog2 + 3log2 - 2log3

Then,

y = log ( [2^(x+3)]/9 )

Converting this into exponential form gives us

10^y = [2^(x+3)]/9

Answer 4

You have to use the basic laws of logs ....
n log m = loh (m^n)
log (mn) = log m + log n
log (m/n) = log m - log n ..... Then

xlog2 + 3log2 - 2log3 = log (2^x * 2^3 / 3^2)

Answer 5

The number in front of the log turns into the exponent:

2^x + 2^3 - 3^2

If you want it simplified:

2^x + 8 - 9
2^x - 1

Answer 6

xlog2 + 3 log2 - 2log3

log 2^x + log2^3 - log3^2

log2^x + log8-log9

log ((8*2^x)/9)=y note: 8 = 2^3, so 2^3*2^x=2^(x+3)

10^y = 2^((x+3)/9)

Answer 7

2^x*2^3*3^-2

Answer 8

=log2^x+log2^3-log3^2
=log2^(x+3)-log9
=log (2^(x+3)/9)