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# logs :(

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Find the solution of the equation ln24x-1 = ln8x+5+ log2161-2x expressing your answer in terms of ln2.

Dec 3, 2019

#1
+1

$$ln2^{4x-1}=ln8^{x+5}+log_2 16^{1-2x}$$

the right handside $$log_2 16^{1-2x}$$ we can make the power (1-2x) next to the log so we could solve for 2^y=16 which will be 4

$$(1-2x)*log_2 16$$

$$log_2 16 = 4$$

4*(1-2x) + $$ln8^{x+5}$$=$$ln2^{4x-1}$$

4-8x+(x+5)*$$ln8$$=(4x-1)*$$ln2$$

$$\frac{4-8x+(x+5)*ln8}{4x-1}=ln2$$

Expressed in terms of ln2

Further simplify could be done by using calculator

ln8=2.07944154

4-8x+(x+5)*2.07944154

4-8x+2.079444154x+10.3972077

4-5.920x+10.3972077

14.3972077-5.920x/4x-1 =ln2

ln2=0.693147

14.4-5.9x=(4x-1)*0.693147

Rest is just algebra and indeed you will find that

indeed you will find x=1.73589087 approx: 1.74

Which works.

I don't think the question wants the x value so just expressed it in ln2

Dec 3, 2019
edited by Guest  Dec 3, 2019
#2
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I.E a much faster way was just (4x-1)*ln2=$$ln8^{x+5}+log_2 16^{1-2x}$$

Divide by (4x-1)

$$ln2$$ = $$\frac{ln8^{x+5}+log_2 16^{1-2x}}{4x-1}$$

Guest Dec 3, 2019
#3
+15
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the answer key says that it should be x=(16ln2+4)/(ln2+8)

i just dont know how to get there

Dec 3, 2019
#4
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But the model answer solved for x, While the question asked for an expression in terms of ln2

Guest Dec 3, 2019
#5
+15
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i think it means solve for x and show what x is in terms of ln2

cherrypiedelicious  Dec 3, 2019
#6
0

ohh! sorry!

I gtg now, i am pretty sure someone will help you soon! (If not i will do later)

Guest Dec 3, 2019
#7
+106993
+1

Find the solution of the equation .... expressing your answer in terms of ln2.

$$ln2^{4x-1} = ln8^{x+5}+ log_216^{1-2x}\\ \text{I will assume that you mean}\\ ln(2^{4x-1}) = ln(8^{x+5})+ log_2(16^{1-2x})\\ (4x-1)ln(2) =(x+5) ln(8)+ (1-2x)log_2(16)\\ (4x-1)ln(2) =(x+5) ln(2^3)+ (1-2x)log_2(2^4)\\ (4x-1)ln(2) =3(x+5) ln(2)+4 (1-2x)log_2(2)\\ (4x-1)ln(2) =3(x+5) ln(2)+4 (1-2x)\\ [(4x-1)-3(x+5)]ln2=4(1-2x)\\ [4x-1-3x-15)]ln2=4(1-2x)\\ [x-16)]ln2=4(1-2x)\\ xln2-16ln2=4-8x\\ xln2+8x=4+16ln2\\ x(ln2+8)=\\ x=\frac{4(1+4ln2)}{ln2+8}\\ x=\frac{16ln2+4}{ln2+8}$$

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Coding:

ln2^{4x-1} = ln8^{x+5}+ log_216^{1-2x}\\
\text{I will assume that you mean}\\
ln(2^{4x-1}) = ln(8^{x+5})+ log_2(16^{1-2x})\\
(4x-1)ln(2) =(x+5) ln(8)+ (1-2x)log_2(16)\\
(4x-1)ln(2) =(x+5) ln(2^3)+ (1-2x)log_2(2^4)\\
(4x-1)ln(2) =3(x+5) ln(2)+4 (1-2x)log_2(2)\\
(4x-1)ln(2) =3(x+5) ln(2)+4 (1-2x)\\
[(4x-1)-3(x+5)]ln2=4(1-2x)\\
[4x-1-3x-15)]ln2=4(1-2x)\\
[x-16)]ln2=4(1-2x)\\
xln2-16ln2=4-8x\\
xln2+8x=4+16ln2\\
x(ln2+8)=\\
x=\frac{4(1+4ln2)}{ln2+8}\\
x=\frac{16ln2+4}{ln2+8}

Dec 3, 2019