advance algebra help?

Question

i have a homework question. i have to solve for x. the problem is 10^x+3 = 6^2x..... so far i got (x+3)log10 = 2xlog6. how do i isolate x?

Answer 1

(x+3)log10 = 2xlog6.
xlog10+3log10-2xlog6=0
x(log10-log36)=-3log10

x=-3log10/((log10-log36)

or

x=-3/(1-log36)

or

x=-3/log (5/18)


Hint:
remember log10=1

Answer 2

I assume you mean 10^(x+3) = 6^(2x)

Well for this problem...

We get, as you said,

(x+3)log 10 = 2x log 6

log 10 = 1

So x+3 = 2x log 6

x - 2xlog6 = -3

factor out an x

x(1-2log(6)) = -3

x(1-log 36) = -3

x(log 10 - log 36) = 3

x ( log(5/18) ) = -3

x = -3 / log(5/18)

If you need a number, plug and chug to get

x = about 5.39

Answer 3

(x+3)log10 = 2xlog6
xlog10 + 3log10 = 2xlog6
xlog10-2xlog6 = -3log10
x(log10-2log6) = -3log10
x = -3log10/(log10-2log6)