How do you solve #\log _{4}x^{3}+\log _{2}\sqrt{x}=8#?

1 Answer
Noah G
Feb 19, 2017

#x = 16#.

Explanation:

The first step is always to rewrite in a common base, using the rules #log_a n = logn/loga# and #logn^a = alogn#.

#logx^3/log4 + logsqrt(x)/log2 = 8#

#logx^3/(log2^2) + logsqrt(x)/(log2^1) = 8#

#(3logx)/(2log2) + (logx^(1/2))/(1log2) = 8#

#(3logx)/(2log2) + (1/2logx)/(log2) = 8#

Put on a common denominator.

#(3logx)/(2log2) + (2(1/2logx))/(2log2) = 8#

#(3logx)/(2log2) + (logx)/(2log2) = 8#

#(3logx)/(log4) + logx/(log4) = 8#

#3log_4x + log_4x = 8#

#log_4 x^3 + log_4 x = 8#

Now use #log_a n + log_a m = log_a(n xx m)#.

#log_4 (x^3x) = 8#

#log_4 x^4 = 8#

Convert to exponential form. If #log_a b = n#, then #a^n = b#.

#x^4 = 4^8#

#x^4 = (4^2)^4#

#x^4 = 16^4#

It is now clear that #x = 16#. The last step is to check whether the answer we just found is actually valid.

#log_4 (16)^3 + log_2 sqrt(16) =^? 8#

#3log_4(16) + log_2 4 =^? 8#

#3(2) + 2 = 8 color(green)(√)#

Hopefully this helps!

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