Question

Solve for x given that %7D%20" id="TexFormula1" title=" {64}^{2x - 1} = {32}^{x + 1} " alt=" {64}^{2x - 1} = {32}^{x + 1} " align="absmiddle" class="latex-formula">
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Answer 1

Answer:

x = $\frac{11}{7}$

Step-by-step explanation:

note that 64 = $2^{6}$ and 32 = $2^{5}$ , thus

$(2^{6}) ^{2x-1}$ = $(2^{5}) ^{x+1}$ , that is

$2^{6(2x-1)}$ = $2^{5(x+1)}$

$2^{12x-6}$ = $2^{5x+5}$

Since the bases on both sides are equal, then equate the exponents

12x - 6 = 5x + 5 ( subtract 5x from both sides )

7x - 6 = 5 ( add 6 to both sides )

7x = 11 ( divide both sides by 7 )

x = $\frac{11}{7}$